UP Board Solutions Class 9 Maths Chapter 4 Algebraic Identities Ex 4.5

Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.5

Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.5 बीजगणितीय सर्वसमिकाऐं

Question 1. निम्न के मान ज्ञात कीजिए।
(a) \( (25)^3 - (75)^3 + (50)^3 \)
(b) \( (0.2)^3 - (0.3)^3 + (0.1)^3 \)
(c) \( (1.5)^3 - (0.9)^3 - (0.6)^3 \)
(d) \( (-12)^3 + 7^3 + 5^3 \)
Answer:
हलः
(a) \( (25)^3 - (75)^3 + (50)^3 \)
\( = (25)^3 - (25 + 50)^3 + (50)^3 \)
\( = (25)^3 - (25)^3 - (50)^3 - 3 \times 25 \times 50(25 + 50) + (50)^3 = -3 \times 25 \times 50 \times 75 = -281250 \)

(b) \( (0.2)^3 - (0.3)^3 + (0.1)^3 \)
\( = (0.2)^3 - (0.2 + 0.1) + (0.1)^3 \)
\( = (0.2)^3 - (0.2)^3 - (0.1)^3 - 3 \times 0.2 \times 0.1 \times (0.2 + 0.1) + (0.1)^3 = -3 \times 0.2 \times 0.1 \times (0.3) = -0.018 \)

(c) \( (1.5)^3 - (0.9)^3 - (0.6)^3 \)
\( = (0.9 + 0.6)^3 - (0.9)^3 - (0.6)^3 \)
\( = (0.9)^3 + (0.6)^3 + 3 \times 0.9 \times 0.6 \times (0.9 + 0.6) - (0.9)^3 - (0.6)^3 = 3 \times 0.9 \times 0.6 \times 1.5 = 2.430 \)

(d) \( (-12)^3 + 7^3 + 5^3 \)
\( \implies 7^3 + (-7 - 5)^3 + 5^3 \)
\( \implies 7^3 + (-7)^3 + (-5)^3 + 3 \times (-7)(-5)(-7 - 5) = 3 \times -7 \times -5(-12) = 105 \times (-12) = -1260 \)
In simple words: These problems involve simplifying cubic expressions using algebraic identities. When \( a+b+c=0 \), then \( a^3+b^3+c^3 = 3abc \). Alternatively, for expressions like \( a^3-b^3+c^3 \), we can substitute \( b = -(a+c) \) and apply the identity.

🎯 Exam Tip: Remember the identity \( a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \). If \( a+b+c=0 \), then \( a^3+b^3+c^3=3abc \). This is a common shortcut for these types of questions.

Question 2. यदि \( x + y + z = 9 \) और \( x^2 + y^2 + z^2 = 35 \), तब \( x^3 + y^3 + z^3 - 3xyz \) के मान ज्ञात कीजिए।
Answer:
हलः
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) ........(1) \)
\( \because x + y + z = 9 \)
वर्ग करने पर,
\( (x + y + z)^2 = 9^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 81 \)
\( 35 + 2(xy + yz + zx) = 81 \)
\( 2(xy + yz + zx) = 81 - 35 = 46 \)
\( xy + yz + zx = \frac{46}{2} = 23 \)
समी० (1) से
\( x^3 + y^3 + z^3 - 3xyz = 9(35 - 23) = 9 \times 12 = 108 \)
In simple words: We used the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \) to find the value of \( (xy+yz+zx) \). Then, we substituted this value along with the given values into the main identity \( x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \) to get the final result.

🎯 Exam Tip: When given \( x+y+z \) and \( x^2+y^2+z^2 \), immediately think of using the square of the sum identity to find \( (xy+yz+zx) \). This value is crucial for computing \( x^3+y^3+z^3-3xyz \).

Question 3. यदि \( x + y + z = 8 \) और \( xy + yz + zx = 26 \), तब \( x^3 + y^3 + z^3 - 3xyz \) के मान ज्ञात कीजिए।
Answer:
हलः
\( \because x + y + z = 8 \)
वर्ग करने पर
\( (x + y + z)^2 = (8)^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 64 \)
\( x^2 + y^2 + z^2 + 2(26) = 64 \)
\( x^2 + y^2 + z^2 = 64 - 52 = 12 \)
\( \because x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = (8)(12 - 26) = (8)(-14) = -112 \)
In simple words: This problem is similar to the previous one. We first used the square of the sum identity to find the value of \( x^2+y^2+z^2 \), and then applied the cubic identity to calculate \( x^3+y^3+z^3-3xyz \).

🎯 Exam Tip: Always write down the given information and the identity you need to use. Break down the problem into smaller steps, such as finding intermediate values like \( x^2+y^2+z^2 \), to avoid errors.

Question 4. निम्न के मान ज्ञात कीजिए।
(i) \( (3x - 4y + 5z)(9x^2 + 16y^2 + 25z^2 + 12xy - 15yz + 20yz) \)
(ii) \( (3x + 2y + 2z)(9x^2 + 4y^2 + 4z^2 -6xy - 4yz - 6xz) \)
(iii) \( (2x - y + 3z)(4x^2 + y^2 + 9z^2 + 2xy + 3yz -6xz) \)
Answer:
हलः
(i) \( (3x - 4y + 5z)(9x^2 + 16y^2 + 25z^2 + 12xy - 15xz + 20yz) \)
\( = (3x)^3 + (-4y)^3 + (5z)^3 - 3[3x \times (-4y) \times 5z] \)
\( = 27x^3 - 64y^3 + 125z^3 + 180xyz \)

(ii) \( (3x + 2y + 2z)(9x^2 + 4y^2 + 4z^2 - 6xy - 4yz - 6xz) \)
\( = (3x)^3 + (2y)^3 + (2z)^3 - 3[3x \times 2y \times 2z] \)
\( = 27x^3 + 8y^3 + 8z^3 - 36xyz \)

(iii) \( (2x - y + 3z)(4x^2 + y^2 - 9z^2 + 2xy + 3yz - 6xz) \)
\( = (2x)^3 + (-y)^3 + (3z)^3 - 3[2x \times (-y) \times 3z] \)
\( = 8x^3 - y^3 + 27z^3 + 18xyz \)
In simple words: These expressions match the expanded form of \( a^3+b^3+c^3-3abc \), which is \( (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \). By identifying 'a', 'b', and 'c' correctly (including signs), we can directly write the result in the simplified cubic form.

🎯 Exam Tip: Recognize the pattern \( (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \). Pay close attention to the signs of the cross-product terms \( ab, bc, ca \) in the second bracket to determine the signs of \( b \) and \( c \).

Ex 4.5 Algebraic Identities विविध प्रश्नावली

Question 1. सर्वसमिका का प्रयोग करके निम्न के मान ज्ञात कीजिए।
(i) \( 103 \times 107 \)
(ii) \( 95 \times 96 \)
(iii) \( 104 \times 96 \)
Answer:
हलः
(i) \( 103 \times 107 = (100 + 3) \times (100 + 7) = (100)^2 + (3 + 7) \times 100 + 3 \times 7 = 10000 + 1000 + 21 = 11021 \)

(ii) \( 95 \times 96 = (100 - 5) \times (100 - 4) = (100)^2 - (5 + 4) \times 100 + 5 \times 4 = 10000 - 900 + 20 = 9120 \)

(iii) \( 104 \times 96 = (100 + 4) \times (100 - 4) = (100)^2 - (4)^2 = 10000 - 16 = 9984 \)
In simple words: These problems use the identities \( (x+a)(x+b) = x^2+(a+b)x+ab \) and \( (a+b)(a-b) = a^2-b^2 \) to simplify multiplication. We express the numbers as sums or differences from a base like 100, making calculations easier.

🎯 Exam Tip: For products close to a power of 10 (e.g., 100), use \( (x+a)(x+b) \) or \( (x+a)(x-a) \) identities. Choose 'x' strategically to simplify the calculation, often 10, 100, or 1000.

Question 2.
Answer:
हलः
\( x + \frac{1}{x} = 6 \)
वर्ग करने पर
\( (x + \frac{1}{x})^2 = (6)^2 \)
\( x^2 + \frac{1}{x^2} + 2x \times \frac{1}{x} = 36 \)
\( x^2 + \frac{1}{x^2} + 2 = 36 \)
\( x^2 + \frac{1}{x^2} = 36 - 2 = 34 \)
पुनः वर्ग करने पर
\( (x^2 + \frac{1}{x^2})^2 = (34)^2 \)
\( x^4 + \frac{1}{x^4} + 2x^2 \times \frac{1}{x^2} = 1156 \)
\( x^4 + \frac{1}{x^4} + 2 = 1156 \)
\( x^4 + \frac{1}{x^4} = 1156 - 2 = 1154 \)
In simple words: We find higher powers of \( x + \frac{1}{x} \) by repeatedly squaring the expression. Starting with \( x + \frac{1}{x} = 6 \), squaring it gives \( x^2 + \frac{1}{x^2} \), and squaring again gives \( x^4 + \frac{1}{x^4} \).

🎯 Exam Tip: Problems involving \( x + \frac{1}{x} \) or \( x - \frac{1}{x} \) often require squaring (or cubing) to find expressions like \( x^2 + \frac{1}{x^2} \) or \( x^3 + \frac{1}{x^3} \). Remember that \( (a+b)^2 = a^2+b^2+2ab \) simplifies nicely to \( a^2+b^2+2 \) when \( b = \frac{1}{a} \).

Question 3. यदि \( x+\frac{1}{x} = 11 \), सिद्ध कीजिए कि \( x^{2}+\frac{1}{x^{2}} = 119 \)
Answer:
हलः
\( x + \frac{1}{x} = 11 \)
वर्ग करने पर
\( (x + \frac{1}{x})^2 = (11)^2 \)
\( x^2 + \frac{1}{x^2} + 2x \times \frac{1}{x} = 121 \)
\( x^2 + \frac{1}{x^2} + 2 = 121 \)
\( x^2 + \frac{1}{x^2} = 121 - 2 = 119 \)
In simple words: To prove the given expression, we start with the initial equation \( x + \frac{1}{x} = 11 \) and square both sides. Using the identity \( (a+b)^2 = a^2+b^2+2ab \), where \( a=x \) and \( b=\frac{1}{x} \), we simplify to directly get \( x^2 + \frac{1}{x^2} \).

🎯 Exam Tip: For 'prove that' questions, always start with the given condition and manipulate it algebraically using relevant identities until you reach the expression to be proved. Show all intermediate steps clearly.

Question 4. यदि \( x^{2}+\frac{1}{x^{2}} = 66 \), सिद्ध कीजिए कि \( x-\frac{1}{x} = \pm 8 \)
Answer:
हलः
\( x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2 \)
\( 66 = (x - \frac{1}{x})^2 + 2 \)
\( (x - \frac{1}{x})^2 = 66 - 2 = 64 \)
\( \implies x - \frac{1}{x} = \sqrt{64} = \pm 8 \)
In simple words: We are given \( x^2 + \frac{1}{x^2} \) and need to find \( x - \frac{1}{x} \). We use the identity \( (a-b)^2 = a^2+b^2-2ab \), specifically \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \). By substituting the given value, we can solve for \( x - \frac{1}{x} \). Remember to consider both positive and negative roots.

🎯 Exam Tip: To relate \( x^2 + \frac{1}{x^2} \) to \( x - \frac{1}{x} \), use the identity \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \). Similarly, for \( x + \frac{1}{x} \), use \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \). These forms are essential for such proofs.

Question 5. यदि \( x = 4, y =3, z = 2 \) तब सिद्ध कीजिए कि \( 4x^2 + y^2 + 25z^2 + 4xy - 10yz - 20xz = 1 \)
Answer:
हलः
L.H.S. \( = 4x^2 + y^2 + 25z^2 + 4xy - 10yz - 20xz \)
\( = (-2x)^2 + (-y)^2 + (5z)^2 + 2(-2x)(-y) + 2(-y)(5z) + 2(5z)(-2x) \)
\( = (-2x - y + 5z)^2 \)
\( = (-2 \times 4 - 3 + 5 \times 2)^2 \)
\( = (-8 - 3 + 10)^2 \)
\( = (-11 + 10)^2 \)
\( = (-1)^2 = 1 = \text{R.H.S.} \)
In simple words: The left-hand side (L.H.S.) expression is a rearrangement of the square of a trinomial \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \). By careful observation, we identify \( a=-2x \), \( b=-y \), and \( c=5z \). Substituting these into the formula and then plugging in the values of \( x, y, z \) proves the equality.

🎯 Exam Tip: When faced with an expression involving squares and cross-products of three variables, try to identify if it matches the expansion of \( (a+b+c)^2 \). Pay close attention to the signs of the cross-product terms to correctly determine the signs of 'a', 'b', and 'c'.

Question 6. यदि \( x + y = 12 \) व \( xy = 27 \), तब सिद्ध कीजिए कि \( x^3 + y^3 = 756 \)
Answer:
हलः
\( \because x + y = 12 \)
घन करने पर \( (x + y)^3 = (12)^3 \)
\( x^3 + y^3 + 3xy(x + y) = 1728 \)
\( x^3 + y^3 + 3 \times 27(12) = 1728 \)
\( x^3 + y^3 + 972 = 1728 \)
\( x^3 + y^3 = 1728 - 972 = 756 \)
In simple words: We are given the sum and product of two variables, \( x+y \) and \( xy \), and need to find the sum of their cubes, \( x^3+y^3 \). We use the identity for the cube of a sum: \( (x+y)^3 = x^3+y^3+3xy(x+y) \). By substituting the given values into this identity, we can directly calculate the required sum of cubes.

🎯 Exam Tip: For problems involving sum/difference and product of two terms, and requiring sum/difference of their cubes, use the identities: \( (a+b)^3 = a^3+b^3+3ab(a+b) \) or \( (a-b)^3 = a^3-b^3-3ab(a-b) \). This avoids finding individual values of 'a' and 'b'.

Question 7. यदि \( x^{2}+\frac{1}{x^{2}} = 83 \), तब सिद्ध कीजिए कि \( x^{3}-\frac{1}{x^{3}} = 756 \)
Answer:
हलः
\( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2x \times \frac{1}{x} \)
\( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \)
\( (x - \frac{1}{x})^2 = 83 - 2 = 81 \)
\( \implies x - \frac{1}{x} = \sqrt{81} = 9 \) (Since \( x^3-\frac{1}{x^3} \) is positive, we take the positive root for \( x - \frac{1}{x} \). If it was negative we would take -9.)
Now, using the identity for difference of cubes:
\( x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 + \frac{1}{x^2} + x \times \frac{1}{x}) \)
\( x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 + \frac{1}{x^2} + 1) \)
\( = (9)(83 + 1) = 9 \times 84 = 756 \)
In simple words: To find \( x^3 - \frac{1}{x^3} \) when \( x^2 + \frac{1}{x^2} \) is known, we first calculate \( x - \frac{1}{x} \) using the identity \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \). Once \( x - \frac{1}{x} \) is found, we apply the difference of cubes identity \( x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 + \frac{1}{x^2} + 1) \).

🎯 Exam Tip: For proving cubic expressions from squared ones, it's a two-step process. First, deduce the linear sum/difference \( (x \pm \frac{1}{x}) \) from \( (x^2 + \frac{1}{x^2}) \). Then, use the appropriate cubic identity \( (a^3 \pm b^3) \).

Question 8. यदि \( x + y + z = 8 \) व \( xy + yz + zx = 20 \), तब सिद्ध कीजिए कि \( x^3 + y^3 + z^3 - 3xyz = 32 \)
Answer:
हलः
\( \because x + y + z = 8 ..........(1) \)
वर्ग करने पर
\( (x + y + z)^2 = (8)^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 64 \)
\( x^2 + y^2 + z^2 = 64 - 2(xy + yz + zx) \)
\( x^2 + y^2 + z^2 = 64 - 2(20) \)
\( x^2 + y^2 + z^2 = 64 - 40 = 24 \)
\( \therefore x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
\( = (8)[24 - (20)] \)
\( = 8[24 - 20] \)
\( = 8 \times 4 = 32 \)
In simple words: We are given \( x+y+z \) and \( xy+yz+zx \). First, we use the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \) to find \( x^2+y^2+z^2 \). Then, we substitute all known values into the identity \( x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \) to prove the result.

🎯 Exam Tip: This problem is a direct application of the cubic identity involving three variables. Ensure accurate calculation of \( x^2+y^2+z^2 \) as an intermediate step. Be careful with signs when substituting into the main identity.

Question 9. सिद्ध कीजिए कि \( 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = [(x - y)^2 + (y - z)^2 + (z-x)^2] \)
Answer:
हलः
L.H.S. \( = 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx \)
\( = x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx \)
\( = (x^2 + y^2 - 2xy) + (y^2 + z^2 - 2yz) + (z^2 + x^2 - 2zx) \)
\( = (x - y)^2 + (y - z)^2 + (z - x)^2 = \text{R.H.S.} \)
In simple words: To prove this identity, we split each squared term on the L.H.S. into two parts (e.g., \( 2x^2 = x^2+x^2 \)). Then, we rearrange the terms to group them into perfect square binomials using the identity \( (a-b)^2 = a^2+b^2-2ab \), thus converting the L.H.S. into the R.H.S.

🎯 Exam Tip: This is a fundamental identity. When seeing \( 2x^2+2y^2+2z^2-2xy-2yz-2zx \), immediately think of splitting the squared terms and forming perfect squares like \( (x-y)^2 \), \( (y-z)^2 \), \( (z-x)^2 \).

Question 10. \( \left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2} \) का विस्तार कीजिए।
Answer:
हलः
\( \left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2} = \left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2\left(\frac{1}{4} a\right)(1) \)
\( = \frac{1}{16} a^2 + \frac{1}{4} b^2 + 1 - \frac{ab}{4} - b + \frac{a}{2} \)
In simple words: This problem requires expanding a trinomial squared expression. We use the identity \( (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx \). Each term is squared, and then twice the product of each pair of terms is added, taking care of the signs.

🎯 Exam Tip: Carefully apply the identity \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \). Be meticulous with fractions and signs during multiplication. A common mistake is forgetting to multiply by 2 for the cross-product terms.

Question 11. विस्तार कीजिए – \( (-2x + 5y - 3z)^2 \)
Answer:
हलः
\( (-2x + 5y - 3z)^2 = (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-2x)(-3z) \)
\( = 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz \)
In simple words: We expand the given trinomial square by identifying \( a = -2x \), \( b = 5y \), and \( c = -3z \). Then, we apply the identity \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \), ensuring that the signs of the terms are correctly handled throughout the expansion.

🎯 Exam Tip: When dealing with negative terms in \( (a+b+c)^2 \), treat them as sums of negative quantities, e.g., \( (-2x + 5y - 3z)^2 = ((-2x) + (5y) + (-3z))^2 \). This helps in correctly applying the identity and managing the signs in the cross-product terms.

Question 12. \( \left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3} \) का मान ज्ञात कीजिए।
Answer:
हलः
Let \( a = \frac{1}{2}, b = \frac{1}{3}, c = -\frac{5}{6} \).
Check if \( a+b+c = 0 \):
\( \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3}{6} + \frac{2}{6} - \frac{5}{6} = \frac{5-5}{6} = 0 \)
Since \( a+b+c=0 \), we can use the identity \( a^3+b^3+c^3 = 3abc \).
\( \left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(-\frac{5}{6}\right)^{3} = 3 \times \frac{1}{2} \times \frac{1}{3} \times (-\frac{5}{6}) \)
\( = -\frac{5}{12} \)
In simple words: This problem becomes simple by recognizing that the sum of the base numbers \( (\frac{1}{2} + \frac{1}{3} - \frac{5}{6}) \) equals zero. When \( a+b+c=0 \), the identity \( a^3+b^3+c^3 = 3abc \) applies. We just multiply 3 by the three terms to get the answer.

🎯 Exam Tip: Always check if the sum of the terms \( (a+b+c) \) is zero when asked to evaluate the sum of their cubes. If it is, apply the identity \( a^3+b^3+c^3 = 3abc \) directly, as it greatly simplifies the calculation.

Ex 4.5 Algebraic Identities बहुविकल्पीय प्रश्न (Multiple Choice Questions)

Question 1. सही विकल्प का चयन कीजिए।
(a) 320
(b) 322
(c) 321
(d) 222
Answer:
हलः
दिया है: \( x + \frac{1}{x} = 3 \)
दोनों पक्षों का वर्ग करने पर,
\( (x + \frac{1}{x})^2 = 3^2 \)
\( \implies x^2 + \frac{1}{x^2} + 2 = 9 \)
\( \implies x^2 + \frac{1}{x^2} = 9 - 2 = 7 \)
अब, दोनों पक्षों का घन करने पर, (for \( x^6 + \frac{1}{x^6} \))
\( (x^2 + \frac{1}{x^2})^3 = 7^3 \)
\( \implies (x^2)^3 + (\frac{1}{x^2})^3 + 3 \times x^2 \times \frac{1}{x^2}(x^2 + \frac{1}{x^2}) = 343 \)
\( \implies x^6 + \frac{1}{x^6} + 3(7) = 343 \)
\( \implies x^6 + \frac{1}{x^6} + 21 = 343 \)
\( \implies x^6 + \frac{1}{x^6} = 343 - 21 = 322 \)
अतः विकल्प (b) सही है।
In simple words: To find \( x^6 + \frac{1}{x^6} \) from \( x + \frac{1}{x} \), we first square the given equation to find \( x^2 + \frac{1}{x^2} \). Then, we cube this new expression \( (x^2 + \frac{1}{x^2})^3 \) to get \( x^6 + \frac{1}{x^6} \), using the identity \( (a+b)^3 = a^3+b^3+3ab(a+b) \).

🎯 Exam Tip: For higher powers like \( x^6 + \frac{1}{x^6} \), it's often easiest to find \( x^2 + \frac{1}{x^2} \) first (by squaring \( x + \frac{1}{x} \)) and then cube the result. Avoid directly cubing \( x + \frac{1}{x} \) multiple times, as it might lead to more complex intermediate steps.

Question 2. यदि \( x - y = 5 \) व \( xy = 12 \), तब \( x^2 + y^2 = \)
(a) 49
(b) 25
(c) 144
(d) इनमें से कोई नहीं
Answer:
हलः
दिया है: \( x - y = 5 \)
दोनों पक्षों का वर्ग करने पर,
\( (x - y)^2 = 5^2 \)
\( \implies x^2 + y^2 - 2xy = 25 \)
\( \implies x^2 + y^2 - 2(12) = 25 \)
\( \implies x^2 + y^2 = 25 + 24 = 49 \)
अतः विकल्प (a) सही है।
In simple words: We use the algebraic identity for the square of a difference, \( (x-y)^2 = x^2+y^2-2xy \). By substituting the given values of \( x-y \) and \( xy \) into this identity, we can directly find the value of \( x^2+y^2 \).

🎯 Exam Tip: This is a basic application of the identity \( (a-b)^2 = a^2+b^2-2ab \). Ensure you substitute the values correctly and perform the arithmetic operations without errors. This identity is fundamental for many algebraic problems.

Question 3.
(a) 100
(b) 127
(c) 10
(d) 12
Answer:
हलः
यह प्रश्न अपूर्ण है। उपलब्ध विकल्पों के आधार पर, यदि प्रश्न \( x - \frac{1}{x} \) का मान पूछ रहा है और किसी पिछले प्रश्न से \( x^2 + \frac{1}{x^2} = 102 \) का मान दिया गया हो, तब:
\( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \)
\( (x - \frac{1}{x})^2 = 102 - 2 \)
\( (x - \frac{1}{x})^2 = 100 \)
वर्गमूल लेने पर,
\( x - \frac{1}{x} = 10 \)
अतः विकल्प (c) सही है।
In simple words: Assuming the question implies finding \( x - \frac{1}{x} \) from \( x^2 + \frac{1}{x^2} = 102 \), we use the identity \( (x-\frac{1}{x})^2 = x^2+\frac{1}{x^2}-2 \). Substituting the given value and taking the square root gives the answer.

🎯 Exam Tip: When a question is incomplete, try to infer the most probable missing part from the context of similar problems or available options. For expressions like \( x^2 + \frac{1}{x^2} \) leading to \( x - \frac{1}{x} \), the identity \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \) is key.

Question 4.
(a) 194
(b) 144
(c) 124
(d) इनमें से कोई नहीं
Answer:
हलः
यह प्रश्न अधूरा है। यदि प्रश्न \( x + \frac{1}{x} \) का मान दिया गया हो और \( x^4 + \frac{1}{x^4} \) का मान ज्ञात करना हो, उदाहरण के लिए यदि \( x + \frac{1}{x} = 4 \) हो, तब:
\( x + \frac{1}{x} = 4 \)
दोनों पक्षों का वर्ग करने पर,
\( (x + \frac{1}{x})^2 = 4^2 \)
\( x^2 + \frac{1}{x^2} + 2 = 16 \)
\( x^2 + \frac{1}{x^2} = 16 - 2 = 14 \)
पुनः वर्ग करने पर,
\( (x^2 + \frac{1}{x^2})^2 = 14^2 \)
\( x^4 + \frac{1}{x^4} + 2 = 196 \)
\( x^4 + \frac{1}{x^4} = 196 - 2 = 194 \)
अतः विकल्प (a) सही है।
In simple words: Assuming the problem intends to find \( x^4 + \frac{1}{x^4} \) from an initial \( x + \frac{1}{x} \) value (e.g., 4), we apply the squaring operation twice. First, square \( x + \frac{1}{x} \) to get \( x^2 + \frac{1}{x^2} \), then square that result to get \( x^4 + \frac{1}{x^4} \).

🎯 Exam Tip: For problems with incomplete information in MCQs, analyze the options to guess the likely question structure. When dealing with powers like \( x^4 + \frac{1}{x^4} \), successive squaring is the standard approach, as it correctly handles the terms. Pay attention to the \( +2 \) or \( -2 \) adjustment in each step.

Question 5.
(a) 15
(b) 105
(c) 25.
(d) 5
Answer:
हलः
यह प्रश्न भी अपूर्ण है। यदि यह सिद्ध करने को कहा गया हो कि \( x + \frac{1}{x} = 5 \) तब \( x^3 + \frac{1}{x^3} \) का मान क्या होगा? या यदि \( x^3 + \frac{1}{x^3} = 110 \) दिया है और \( x + \frac{1}{x} \) का मान पूछा है। यदि \( x^3 + \frac{1}{x^3} = 110 \) दिया है तो:
\( (x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3x \times \frac{1}{x}(x + \frac{1}{x}) \)
\( (x + \frac{1}{x})^3 = 110 + 3(x + \frac{1}{x}) \)
माना \( y = x + \frac{1}{x} \)
\( y^3 = 110 + 3y \)
\( y^3 - 3y - 110 = 0 \)
तब \( x + \frac{1}{x} = 5 \) रखने पर सर्वसमिका सन्तुष्ट होती है।
\( 5^3 - 3(5) - 110 = 125 - 15 - 110 = 110 - 110 = 0 \)
इसलिए, \( x + \frac{1}{x} = 5 \).
अतः विकल्प (d) सही है।
In simple words: Assuming the question asks to find \( x + \frac{1}{x} \) given \( x^3 + \frac{1}{x^3} = 110 \), we use the identity \( (x+\frac{1}{x})^3 = x^3+\frac{1}{x^3}+3(x+\frac{1}{x}) \). By substituting \( y = x + \frac{1}{x} \), we form a cubic equation \( y^3-3y-110=0 \). Testing the options or factoring shows that \( y=5 \) is a solution.

🎯 Exam Tip: When solving for \( (x \pm \frac{1}{x}) \) from \( (x^3 \pm \frac{1}{x^3}) \), set \( y = (x \pm \frac{1}{x}) \) and form a cubic equation in 'y'. Then, either factor the cubic polynomial or test the given options to find the correct value. This method is efficient for MCQs.

Question 6.
(a) 17
(b) 4
(c) 17/4
(d) 19/4
Answer:
हलः
यह प्रश्न अपूर्ण है। यदि प्रश्न \( x - \frac{1}{x} = \frac{15}{4} \) दिया गया हो और \( x + \frac{1}{x} \) का मान ज्ञात करना हो, तब:
हम जानते हैं कि \( (x + \frac{1}{x})^2 = (x - \frac{1}{x})^2 + 4 \)
\( (x + \frac{1}{x})^2 = (\frac{15}{4})^2 + 4 \)
\( (x + \frac{1}{x})^2 = \frac{225}{16} + 4 \)
\( (x + \frac{1}{x})^2 = \frac{225 + 64}{16} = \frac{289}{16} \)
वर्गमूल लेने पर,
\( x + \frac{1}{x} = \sqrt{\frac{289}{16}} = \frac{17}{4} \)
अतः विकल्प (c) सही है।
In simple words: Assuming the question provides \( x - \frac{1}{x} = \frac{15}{4} \) and asks for \( x + \frac{1}{x} \), we use the identity \( (x+ \frac{1}{x})^2 = (x- \frac{1}{x})^2 + 4 \). Substitute the given value of \( x - \frac{1}{x} \), calculate the square, add 4, and then take the square root to find \( x + \frac{1}{x} \).

🎯 Exam Tip: Remember the crucial relationship between \( (a+b)^2 \) and \( (a-b)^2 \): \( (a+b)^2 = (a-b)^2 + 4ab \). When \( b = \frac{1}{a} \), this simplifies to \( (x+\frac{1}{x})^2 = (x-\frac{1}{x})^2 + 4 \). This identity is frequently tested.

Question 7. यदि \( x + y + z = 9, xy + yz + xz = 23 \), तब \( x^3 + y^3 + z^3 - 3xyz = \)
(a) 100
(b) 81
(c) 108
(d) 123
Answer:
हलः
दिया है: \( x + y + z = 9 \)
दोनों पक्षों का वर्ग करने पर,
\( (x + y + z)^2 = 81 \)
\( \implies x^2 + y^2 + z^2 + 2(xy + yz + zx) = 81 \)
\( \implies x^2 + y^2 + z^2 = 81 - 2 \times 23 = 81 - 46 = 35 \)
\( \therefore x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
\( = 9 \times (35 - 23) \)
\( = 9 \times 12 = 108 \)
अतः विकल्प (c) सही है।
In simple words: First, we use the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \) to calculate \( x^2+y^2+z^2 \). Then, we substitute all the known values (\( x+y+z \), \( x^2+y^2+z^2 \), and \( xy+yz+zx \)) into the cubic identity \( x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \) to find the result.

🎯 Exam Tip: This is a standard problem involving the identity \( a^3+b^3+c^3-3abc \). Always make sure to calculate \( a^2+b^2+c^2 \) correctly as an intermediate step. Double-check your arithmetic, especially when dealing with subtractions inside the final bracket.

Question 8. यदि \( \frac{x}{y}+\frac{y}{x} = 1 \), तब \( x^3 + y^3 = \)
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
हलः
दिया है: \( \frac{x}{y}+\frac{y}{x} = 1 \)
\( \implies \frac{x^2+y^2}{xy} = 1 \)
\( \implies x^2+y^2 = xy \)
\( \implies x^2+y^2-xy = 0 \)
दोनों पक्षों का घन करने पर, (या \( x^3+y^3 = (x+y)(x^2-xy+y^2) \) का प्रयोग करके)
\( x^3+y^3 = (x+y)(x^2-xy+y^2) \)
\( x^3+y^3 = (x+y)(0) \)
\( x^3+y^3 = 0 \)
अतः विकल्प (d) सही है।
In simple words: By simplifying the given equation \( \frac{x}{y}+\frac{y}{x} = 1 \), we find that \( x^2+y^2-xy = 0 \). Knowing this, we use the identity for the sum of cubes, \( x^3+y^3 = (x+y)(x^2-xy+y^2) \). Substituting \( x^2+y^2-xy = 0 \) directly yields \( x^3+y^3=0 \).

🎯 Exam Tip: When you see expressions like \( \frac{x}{y}+\frac{y}{x} = 1 \), immediately simplify it to \( x^2+y^2=xy \) or \( x^2-xy+y^2=0 \). This identity is crucial and often leads to \( x^3+y^3=0 \), as \( x^3+y^3 = (x+y)(x^2-xy+y^2) \).

Question 9. यदि \( x - y = -8 \) व \( xy = -12 \), तब \( x^3 - y^3 = \)
(a) 224
(b) -224
(c) 234
(d) -234
Answer:
हलः
दिया है: \( x - y = -8 \)
दोनों पक्षों का घन करने पर,
\( (x - y)^3 = (-8)^3 \)
\( \implies x^3 - y^3 - 3xy(x - y) = -512 \)
\( \implies x^3 - y^3 = -512 + 3xy(x - y) \)
\( \implies x^3 - y^3 = -512 + 3(-12)(-8) \)
\( \implies x^3 - y^3 = -512 + 288 \)
\( \implies x^3 - y^3 = -224 \)
अतः विकल्प (b) सही है।
In simple words: We use the identity for the cube of a difference: \( (x-y)^3 = x^3-y^3-3xy(x-y) \). By substituting the given values of \( x-y \) and \( xy \) into this identity, we directly calculate the value of \( x^3-y^3 \). Be careful with the negative signs in the calculation.

🎯 Exam Tip: This problem is a direct application of the identity \( (a-b)^3 = a^3-b^3-3ab(a-b) \). The key is to correctly substitute the negative values for \( x-y \) and \( xy \) and perform the arithmetic accurately to get the final result.

Question 10. यदि \( x + y + z = 9 \) व \( xy + yz + zx = 23 \), तब \( x^2 + y^2 + z^2 = \)
(a) 25
(b) 35
(c) 45
(d) 305
Answer:
हलः
दिया है: \( x + y + z = 9 \)
दोनों पक्षों का वर्ग करने पर,
\( (x + y + z)^2 = 9^2 \)
\( \implies x^2 + y^2 + z^2 + 2(xy + yz + zx) = 81 \)
\( \implies x^2 + y^2 + z^2 = 81 - 2(xy + yz + zx) \)
\( \implies x^2 + y^2 + z^2 = 81 - 2 \times 23 \) ( \( \because xy + yz + zx = 23 \) )
\( \implies x^2 + y^2 + z^2 = 81 - 46 = 35 \)
अतः विकल्प (b) सही है।
In simple words: We are given the sum of three variables and the sum of their pairwise products. We use the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \) to find \( x^2+y^2+z^2 \). By rearranging the formula and substituting the given values, we can directly solve for the required sum of squares.

🎯 Exam Tip: This is a straightforward application of the square of a trinomial identity. Ensure you correctly rearrange the formula to isolate \( x^2+y^2+z^2 \) and perform the subtraction accurately. This identity is fundamental for problems involving three variables.

Ex 4.5 Algebraic Identities स्वमूल्यांकन परीक्षण (Self Assessment Test)

Question 1. \( (2x – y + z)^2 \) का विस्तार कीजिए ।
Answer:
हलः
\( (2x - y + z)^2 = (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z) \)
\( = 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz \)
In simple words: To expand \( (2x - y + z)^2 \), we apply the identity \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \). We identify \( a=2x \), \( b=-y \), and \( c=z \), then substitute these into the formula and simplify all terms, being careful with signs.

🎯 Exam Tip: When expanding trinomials, carefully assign the correct signs to 'a', 'b', and 'c' (e.g., \( -y \) as \( (-y) \)). Then apply the standard formula, multiplying each cross-product by 2. Accuracy in signs and coefficients is key.

Question 2. \( (3a + 4b + 5c)^2 \) का विस्तार कीजिए।
Answer:
हलः
\( (3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a) \)
\( = 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ac \)
In simple words: This problem involves expanding a trinomial where all terms are positive. We apply the identity \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \). Square each term individually, and then add twice the product of every possible pair of terms.

🎯 Exam Tip: For expansions like this, ensure you correctly square each term and correctly calculate each of the three pairwise products, multiplying each by 2. This is a direct application of a fundamental identity.

Question 3. \( (4a – 2b – 3c)^2 \) का विस्तार कीजिए।
Answer:
हलः
\( (4a - 2b - 3c)^2 = (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a) \)
\( = 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ac \)
In simple words: To expand \( (4a - 2b - 3c)^2 \), we treat the expression as \( (4a + (-2b) + (-3c))^2 \) and use the identity \( (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx \). Carefully square each term and compute twice the product of each pair, paying close attention to how the negative signs multiply.

🎯 Exam Tip: When terms are negative, ensure their squares become positive (e.g., \( (-2b)^2 = 4b^2 \)). In the cross-product terms, apply standard multiplication rules for signs (e.g., \( 2(-2b)(-3c) = +12bc \)). This is where most errors occur.

Question 4. सर्वसमिका का प्रयोग करके \( (28)^3 + (-15)^3 + (-13)^3 \) का मान ज्ञात कीजिए।
Answer:
हलः
दिया है: \( a = 28, b = -15, c = -13 \)
पहले योग जाँच करें: \( a+b+c = 28 + (-15) + (-13) = 28 - 15 - 13 = 28 - 28 = 0 \)
चूँकि \( a+b+c = 0 \), हम सर्वसमिका \( a^3+b^3+c^3 = 3abc \) का प्रयोग कर सकते हैं।
\( (28)^3 + (-15)^3 + (-13)^3 = 3 \times 28 \times (-15) \times (-13) \)
\( = 3 \times 28 \times (15 \times 13) \)
\( = 3 \times 28 \times 195 \)
\( = 84 \times 195 \)
\( = 16380 \)
In simple words: We check if the sum of the numbers, \( 28 + (-15) + (-13) \), is zero. Since it is, we apply the identity \( a^3+b^3+c^3 = 3abc \). This simplifies the calculation to just multiplying \( 3 \times 28 \times (-15) \times (-13) \).

🎯 Exam Tip: For problems asking for the sum of cubes of three numbers, always verify if their sum is zero. This is a common trick, and if \( a+b+c=0 \), then \( a^3+b^3+c^3 = 3abc \), which drastically simplifies the problem. Don't forget the negative signs in multiplication.

Question 5. \( (104)^3 \) का मान सर्वसमिका का प्रयोग करके ज्ञात कीजिए ।
Answer:
हलः
\( (104)^3 = (100 + 4)^3 \)
सर्वसमिका \( (a+b)^3 = a^3+b^3+3ab(a+b) \) का प्रयोग करने पर:
\( = (100)^3 + (4)^3 + 3 \times 100 \times 4(100 + 4) \)
\( = 1000000 + 64 + 1200(104) \)
\( = 1000000 + 64 + 124800 \)
\( = 1124864 \)
In simple words: To calculate \( (104)^3 \) using an identity, we express 104 as \( (100+4) \). Then, we apply the cubic identity \( (a+b)^3 = a^3+b^3+3ab(a+b) \), where \( a=100 \) and \( b=4 \). We calculate each term and sum them up for the final result.

🎯 Exam Tip: For cubing numbers close to multiples of 10 (like 100), express them as \( (x+a)^3 \) or \( (x-a)^3 \). Applying the respective identities \( (a+b)^3 = a^3+b^3+3ab(a+b) \) or \( (a-b)^3 = a^3-b^3-3ab(a-b) \) makes the calculation easier than direct multiplication.

Question 6. यदि \( 3x – 7y = 10 \) व \( xy = -1 \), तब सिद्ध कीजिए कि \( 9x^2 + 49y^2 = 58 \)
Answer:
हलः
दिया है: \( 3x - 7y = 10 \)
दोनों ओर का वर्ग करने पर
\( (3x - 7y)^2 = (10)^2 \)
\( (3x)^2 + (7y)^2 - 2(3x)(7y) = 100 \)
\( 9x^2 + 49y^2 - 42xy = 100 \)
\( 9x^2 + 49y^2 - 42(-1) = 100 \)
\( 9x^2 + 49y^2 + 42 = 100 \)
\( 9x^2 + 49y^2 = 100 - 42 = 58 \)
In simple words: We are given \( 3x-7y=10 \) and \( xy=-1 \). To prove \( 9x^2+49y^2=58 \), we square the expression \( (3x-7y) \). This expands to \( (3x)^2 + (7y)^2 - 2(3x)(7y) \), which simplifies to \( 9x^2+49y^2-42xy \). Substituting the value of \( xy \) then leads to the desired result.

🎯 Exam Tip: When asked to prove a sum of squares from a difference and product of terms, square the difference. The identity \( (a-b)^2 = a^2+b^2-2ab \) is essential. Be careful with signs, especially when \( xy \) is negative.

Question 7. सिद्ध कीजिए कि \( x^2 + y^2 + z^2 – xy – yz – zx; x, y \) व \( z \) के सभी मानों के लिए सदैव धनात्मक होगा।
Answer:
हलः
हम जानते हैं कि
\( x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2}[(x - y)^2 + (y - z)^2 + (z - x)^2] \)
\( \because (x - y)^2, (y - z)^2, (z - x)^2 \) पूर्ण वर्ग हैं जो हमेशा धनात्मक होते हैं।
अर्थात्, \( (x - y)^2 \ge 0, (y - z)^2 \ge 0, (z - x)^2 \ge 0 \)
अतः इसका योग सदैव धनात्मक होगा।
\( \implies (x - y)^2 + (y - z)^2 + (z - x)^2 \ge 0 \)
\( \implies \frac{1}{2}[(x - y)^2 + (y - z)^2 + (z - x)^2] \ge 0 \)
इसलिए, \( x^2 + y^2 + z^2 - xy - yz - zx \) सदैव धनात्मक होगा।
In simple words: The expression \( x^2+y^2+z^2-xy-yz-zx \) can be rewritten as \( \frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2] \). Since any real number squared is non-negative, the sum of three squared terms must also be non-negative. Therefore, the original expression is always positive or zero.

🎯 Exam Tip: This is a key identity to remember: \( x^2+y^2+z^2-xy-yz-zx = \frac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2] \). Knowing that squares of real numbers are always non-negative is fundamental for proving such expressions are always positive.

Question 8. यदि \( x^2 + y^2 + z^2 = 20 \) व \( x + y + z = 0 \), तब सिद्ध कीजिए \( xy + yz + zx = -10 \)
Answer:
हलः
दिया है: \( x + y + z = 0 \)
वर्ग करने पर, \( (x + y + z)^2 = 0^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 0 \)
\( 20 + 2(xy + yz + zx) = 0 \)
\( 2(xy + yz + zx) = 0 - 20 \)
\( 2(xy + yz + zx) = -20 \)
\( xy + yz + zx = \frac{-20}{2} = -10 \)
In simple words: We are given \( x+y+z=0 \) and \( x^2+y^2+z^2=20 \). To prove \( xy+yz+zx=-10 \), we square the expression \( (x+y+z) \). Using the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \) and substituting the given values, we can directly solve for \( xy+yz+zx \).

🎯 Exam Tip: When \( x+y+z=0 \), the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \) simplifies to \( 0 = x^2+y^2+z^2+2(xy+yz+zx) \). This makes it easy to find \( xy+yz+zx \) if \( x^2+y^2+z^2 \) is known, or vice versa.

Question 9. यदि \( x + y + z = 6 \) व \( xy + yz + zx = 11 \), तब सिद्ध कीजिए \( x^3 + y^3 + z^3 - 3xyz = 18 \)
Answer:
हलः
\( \because x + y + z = 6 ................ (1) \)
हम जानते हैं कि: \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) ........(2) \)
समी० (1) का वर्ग करने पर
\( (x + y + z)^2 = 6^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 36 \)
\( x^2 + y^2 + z^2 + 2(11) = 36 \)
\( x^2 + y^2 + z^2 = 36 - 22 = 14 \)
समी॰ (2) में \( x^2 + y^2 + z^2 \) का मान रखने पर
\( x^3 + y^3 + z^3 - 3xyz = (6) (14 - 11) \)
\( = (6) (3) = 18 \)
In simple words: To prove the cubic identity, we first find \( x^2+y^2+z^2 \) by squaring \( (x+y+z) \) and using the given value of \( (xy+yz+zx) \). Then, we substitute all these values into the main identity \( x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \) to get the result.

🎯 Exam Tip: This problem reinforces the multi-step application of algebraic identities. Remember to always calculate \( x^2+y^2+z^2 \) as an intermediate value using the square of the sum identity before applying the cubic identity.

Question 10. \( x^3 + y^3 + z^3 \) का मान ज्ञात कीजिए, यदि \( x + y + z = 11, x^2 + y^2 + z^2 = 45 \) व \( xyz = 40 \)
Answer:
हलः
दिया है: \( x + y + z = 11 ..............(1) \)
समी० (1) का वर्ग करने पर
\( \therefore (x + y + z)^2 = (11)^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 121 \)
\( 45 + 2(xy + yz + zx) = 121 \)
\( 2(xy + yz + zx) = 121 - 45 = 76 \)
\( \therefore xy + yz + zx = \frac{76}{2} = 38 \)
अब, हम जानते हैं कि:
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
\( x^3 + y^3 + z^3 - 3(40) = (11)(45 - 38) \)
\( x^3 + y^3 + z^3 - 120 = (11)(7) \)
\( x^3 + y^3 + z^3 - 120 = 77 \)
\( x^3 + y^3 + z^3 = 77 + 120 = 197 \)
In simple words: First, we find \( xy+yz+zx \) using the given \( x+y+z \) and \( x^2+y^2+z^2 \) in the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \). Then, we substitute all known values, including \( xyz \), into the identity \( x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \) and solve for \( x^3+y^3+z^3 \).

🎯 Exam Tip: This question requires a complete understanding and application of both the square of the sum and the cubic sum identities for three variables. Systematically calculate intermediate values like \( xy+yz+zx \) before plugging them into the final identity. Always isolate the required term \( x^3+y^3+z^3 \) in the final step.

Question 11. \( x^3 + y^3 + z^3 \) का मान ज्ञात कीजिए, यदि \( x + y + z = 15, xy + yz + zx = 71 \) व \( xyz = 10 \)
Answer:
हलः
हम जानते हैं कि:
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
यह समीकरण (1) है। हमें \( x^2 + y^2 + z^2 \) का मान ज्ञात करने की आवश्यकता है।
दिया है: \( x + y + z = 15 \)....(2)
समी० (2) का वर्ग करने पर
\( (x + y + z)^2 = 15^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 225 \)
\( x^2 + y^2 + z^2 + 2(71) = 225 \)
\( x^2 + y^2 + z^2 = 225 - 142 \)
\( x^2 + y^2 + z^2 = 83 \)
अब \( x^2 + y^2 + z^2 = 83 \) का मान समीकरण (1) में रखने पर,
\( x^3 + y^3 + z^3 - 3(10) = (15)(83 - 71) \)
\( x^3 + y^3 + z^3 - 30 = (15)(12) \)
\( x^3 + y^3 + z^3 - 30 = 180 \)
\( x^3 + y^3 + z^3 = 180 + 30 = 210 \)
In simple words: This problem is similar to the previous one. We first calculate \( x^2+y^2+z^2 \) using the given \( x+y+z \) and \( xy+yz+zx \) from the identity for the square of a sum. Then, we plug all the obtained values, including \( xyz \), into the cubic identity to find \( x^3+y^3+z^3 \).

🎯 Exam Tip: Consistent application of identities is crucial. Always prioritize finding the missing components (like \( x^2+y^2+z^2 \)) that are needed for the main identity. Careful substitution and arithmetic are key to achieving the correct answer.

Question 12. \( x^3 – 8y^3 – 36xy – 216 \) का मान ज्ञात कीजिए, यदि \( x = 2y + 6 \)
Answer:
हलः
दिया है: \( x = 2y + 6 \)
इसे हम लिख सकते हैं: \( x - 2y - 6 = 0 \)
माना \( a = x, b = -2y, c = -6 \)
तो \( a+b+c = x - 2y - 6 = 0 \)
जब \( a+b+c=0 \), तब \( a^3+b^3+c^3 = 3abc \) होता है।
इसलिए, \( x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6) \)
\( x^3 - 8y^3 - 216 = 36xy \)
\( x^3 - 8y^3 - 36xy - 216 = 0 \)
अतः, दिए गए व्यंजक का मान 0 है।
(Alternatively, by direct substitution):
\( x = 2y + 6 \) का मान रखने पर,
\( (2y + 6)^3 - 8y^3 - 36(2y + 6)y - 216 \)
\( = (2y)^3 + (6)^3 + 3(2y)(6)(2y + 6) - 8y^3 - 36y(2y + 6) - 216 \)
\( = 8y^3 + 216 + 36y(2y + 6) - 8y^3 - 72y^2 - 216y - 216 \)
\( = 8y^3 + 216 + 72y^2 + 216y - 8y^3 - 72y^2 - 216y - 216 \)
\( = 0 \)
In simple words: The problem provides a condition \( x=2y+6 \), which can be rewritten as \( x-2y-6=0 \). This means the sum of three terms (\( x \), \( -2y \), \( -6 \)) is zero. For three terms whose sum is zero, the sum of their cubes is \( 3 \) times their product: \( x^3+(-2y)^3+(-6)^3 = 3(x)(-2y)(-6) \). Rearranging this gives \( x^3-8y^3-216 = 36xy \), which means the given expression \( x^3-8y^3-36xy-216 \) equals zero.

🎯 Exam Tip: When simplifying an expression and a condition is given (e.g., \( x=2y+6 \)), try to manipulate the condition into the form \( a+b+c=0 \). If successful, apply the identity \( a^3+b^3+c^3=3abc \) directly, as it significantly reduces calculation complexity. Direct substitution is also a valid, but often longer, alternative.

Question 13. \( x^3 + y^3 + z^3 - 3xyz \) का मान ज्ञात कीजिए, यदि \( x + y + z = 14 \) व \( x^2 + y^2 + z^2 = 60 \)
Answer:
हलः
दिया है: \( x + y + z = 14 \).................... .(1)
वर्ग करने पर,
\( (x + y + z)^2 = 14^2 \)
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 196 \)
\( 60 + 2(xy + yz + zx) = 196 \)
\( 2(xy + yz + zx) = 196 - 60 = 136 \)
\( xy + yz + zx = \frac{136}{2} = 68 \)
अब, \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
\( = (14)(60 - 68) \)
\( = 14 \times (-8) \)
\( = -112 \)
In simple words: We are given the sum of three variables and the sum of their squares. First, we calculate the sum of their pairwise products, \( xy+yz+zx \), using the identity \( (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \). Then, we substitute all these values into the cubic identity \( x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \) to find the result.

🎯 Exam Tip: This is a multi-step problem. Always calculate \( (xy+yz+zx) \) first if not directly provided. Ensure careful substitution into the main cubic identity, especially with negative results from subtractions, to avoid sign errors in the final answer.

Question 14. यदि \( 4x^2 + y^2 = 40 \) व \( xy =6 \), तब सिद्ध कीजिए कि \( 2x + y = \pm 8 \)
Answer:
हलः
दिया है: \( 4x^2 + y^2 = 40 \)
हम जानते हैं कि \( (2x + y)^2 = (2x)^2 + y^2 + 2(2x)(y) \)
\( (2x + y)^2 = 4x^2 + y^2 + 4xy \)
दिए गए मानों को रखने पर,
\( (2x + y)^2 = 40 + 4(6) \)
\( (2x + y)^2 = 40 + 24 \)
\( (2x + y)^2 = 64 \)
\( \implies 2x + y = \pm \sqrt{64} = \pm 8 \)
In simple words: To prove \( 2x+y = \pm 8 \), we start by considering the square of \( (2x+y) \). We use the identity \( (a+b)^2 = a^2+b^2+2ab \), where \( a=2x \) and \( b=y \). Substituting the given values \( 4x^2+y^2=40 \) and \( xy=6 \) into the expanded form allows us to find \( (2x+y)^2 \) and then take its square root.

🎯 Exam Tip: This problem connects squared terms with linear sums. Recognizing that \( 4x^2 \) is \( (2x)^2 \) is key. Then, directly apply the \( (a+b)^2 \) identity, substitute the given values, and remember to include both positive and negative roots when taking the square root.

Question 15.
Answer:
हलः
यह प्रश्न भी अपूर्ण है। यदि प्रश्न में \( x - \frac{1}{x} = -1 \) दिया है और \( x^4 + \frac{1}{x^4} \) का मान पूछा गया हो, तब:
दिया है: \( x - \frac{1}{x} = -1 \)
वर्ग करने पर,
\( (x - \frac{1}{x})^2 = (-1)^2 \)
\( x^2 + \frac{1}{x^2} - 2x \times \frac{1}{x} = 1 \)
\( \implies x^2 + \frac{1}{x^2} - 2 = 1 \)
\( \implies x^2 + \frac{1}{x^2} = 1 + 2 = 3 \)
पुनः वर्ग करने पर,
\( (x^2 + \frac{1}{x^2})^2 = 3^2 \)
\( x^4 + \frac{1}{x^4} + 2x^2 \times \frac{1}{x^2} = 9 \)
\( \implies x^4 + \frac{1}{x^4} + 2 = 9 \)
\( \implies x^4 + \frac{1}{x^4} = 9 - 2 = 7 \)
In simple words: Assuming we need to find \( x^4+\frac{1}{x^4} \) from \( x-\frac{1}{x}=-1 \), we square the given equation twice. First, squaring \( x-\frac{1}{x} \) gives \( x^2+\frac{1}{x^2} \). Then, squaring \( x^2+\frac{1}{x^2} \) gives \( x^4+\frac{1}{x^4} \). Each step involves using the identity \( (a \pm b)^2 = a^2+b^2 \pm 2ab \).

🎯 Exam Tip: For problems involving higher powers of \( x \pm \frac{1}{x} \), such as \( x^4 + \frac{1}{x^4} \), perform successive squaring. Be mindful of the \( -2 \) term when starting with \( x - \frac{1}{x} \) and the \( +2 \) term in subsequent steps for \( x^n + \frac{1}{x^n} \).

Question 16. यदि \( 2x + 3y = 8 \) व \( xy = 2 \), तब सिद्ध कीजिए कि \( 4x^2 + 9y^2 = 40 \)
Answer:
हलः
दिया है: \( 2x + 3y = 8 \)
वर्ग करने पर, \( (2x + 3y)^2 = 8^2 \)
\( (2x)^2 + (3y)^2 + 2(2x)(3y) = 64 \)
\( 4x^2 + 9y^2 + 12xy = 64 \)
\( 4x^2 + 9y^2 + 12(2) = 64 \)
\( 4x^2 + 9y^2 + 24 = 64 \)
\( 4x^2 + 9y^2 = 64 - 24 \)
\( 4x^2 + 9y^2 = 40 \)
In simple words: To prove \( 4x^2+9y^2=40 \), we take the given linear equation \( 2x+3y=8 \) and square both sides. Using the identity \( (a+b)^2 = a^2+b^2+2ab \), where \( a=2x \) and \( b=3y \), we expand and substitute the value of \( xy \) to directly obtain the desired expression.

🎯 Exam Tip: This is a direct application of squaring a binomial sum. The key is to correctly identify \( a=2x \) and \( b=3y \), then apply \( (a+b)^2 = a^2+b^2+2ab \). Carefully substitute the given product term \( xy \) and perform arithmetic to reach the proof.

Question 17.
Answer:
हलः
यह प्रश्न भी अपूर्ण है। यदि \( x^4 + \frac{1}{x^4} = 47 \) दिया है, तो \( x + \frac{1}{x} \) का मान ज्ञात कीजिए।
दिया है: \( x^4 + \frac{1}{x^4} = 47 \)
हम जानते हैं कि \( (x^2 + \frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2 \)
\( (x^2 + \frac{1}{x^2})^2 = 47 + 2 = 49 \)
\( \implies x^2 + \frac{1}{x^2} = \sqrt{49} = 7 \) (धनात्मक मान लेते हैं क्योंकि \( x^2+\frac{1}{x^2} \) हमेशा धनात्मक होगा)
पुनः, हम जानते हैं कि \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \)
\( (x + \frac{1}{x})^2 = 7 + 2 = 9 \)
\( \implies x + \frac{1}{x} = \sqrt{9} = 3 \) (धनात्मक मान लेते हैं यदि \( x > 0 \) है)
In simple words: Assuming we need to find \( x+\frac{1}{x} \) from \( x^4+\frac{1}{x^4}=47 \), we reverse the squaring process. First, we find \( x^2+\frac{1}{x^2} \) using \( (x^2+\frac{1}{x^2})^2 = x^4+\frac{1}{x^4}+2 \). Then, we find \( x+\frac{1}{x} \) using \( (x+\frac{1}{x})^2 = x^2+\frac{1}{x^2}+2 \). We take positive square roots at each step.

🎯 Exam Tip: To find lower powers (e.g., \( x+\frac{1}{x} \)) from higher even powers (e.g., \( x^4+\frac{1}{x^4} \)), apply the reverse squaring method. Remember that \( A^2 = B+2 \implies A=\sqrt{B+2} \). Also, take appropriate square roots (positive or negative) based on the context of the problem.

Question 18.
Answer:
हलः
यह प्रश्न अपूर्ण है। यदि \( x - \frac{1}{x} = 7 \) दिया है और \( x^3 - \frac{1}{x^3} \) का मान ज्ञात करना हो, तब:
दिया है: \( x - \frac{1}{x} = 7 \)
दोनों पक्षों का घन करने पर,
\( (x - \frac{1}{x})^3 = 7^3 \)
\( x^3 - \frac{1}{x^3} - 3x \times \frac{1}{x}(x - \frac{1}{x}) = 343 \)
\( x^3 - \frac{1}{x^3} - 3(7) = 343 \)
\( x^3 - \frac{1}{x^3} - 21 = 343 \)
\( x^3 - \frac{1}{x^3} = 343 + 21 \)
\( x^3 - \frac{1}{x^3} = 364 \)
In simple words: Assuming we need to calculate \( x^3 - \frac{1}{x^3} \) given \( x - \frac{1}{x} = 7 \), we cube both sides of the equation. Using the identity \( (a-b)^3 = a^3-b^3-3ab(a-b) \), we substitute \( a=x \) and \( b=\frac{1}{x} \) to directly find the value of \( x^3 - \frac{1}{x^3} \).

🎯 Exam Tip: To find \( x^3 \pm \frac{1}{x^3} \) from \( x \pm \frac{1}{x} \), directly cube the given linear expression. Remember the simplified identities: \( (x-\frac{1}{x})^3 = x^3-\frac{1}{x^3}-3(x-\frac{1}{x}) \) and \( (x+\frac{1}{x})^3 = x^3+\frac{1}{x^3}+3(x+\frac{1}{x}) \).

Question 19. सिद्ध कीजिए कि \( (x + y + z)^2 – (x – y – z)^2 = 4x (y + z) \)
Answer:
हलः
L.H.S. \( = (x + y + z)^2 – (x – y – z)^2 \)
Let \( A = (y+z) \). Then the expression becomes \( (x+A)^2 - (x-A)^2 \).
Using the identity \( (a+b)^2 - (a-b)^2 = 4ab \), where \( a=x \) and \( b=A \):
\( = 4x(y+z) \)
Alternatively, by expanding:
L.H.S. \( = (x^2 + y^2 + z^2 + 2xy + 2yz + 2zx) – (x^2 + (-y)^2 + (-z)^2 + 2x(-y) + 2(-y)(-z) + 2x(-z)) \)
\( = (x^2 + y^2 + z^2 + 2xy + 2yz + 2zx) – (x^2 + y^2 + z^2 - 2xy + 2yz - 2zx) \)
\( = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx - x^2 - y^2 - z^2 + 2xy - 2yz + 2zx \)
\( = (x^2 - x^2) + (y^2 - y^2) + (z^2 - z^2) + (2xy + 2xy) + (2yz - 2yz) + (2zx + 2zx) \)
\( = 0 + 0 + 0 + 4xy + 0 + 4zx \)
\( = 4xy + 4xz \)
\( = 4x(y + z) = \text{R.H.S.} \)
In simple words: This identity can be proven by treating \( (y+z) \) as a single term, say 'A'. The expression then becomes \( (x+A)^2 - (x-A)^2 \). Using the identity \( (a+b)^2 - (a-b)^2 = 4ab \), we directly get \( 4x(y+z) \). Alternatively, expanding both squares and simplifying also yields the same result.

🎯 Exam Tip: Recognizing the pattern \( (A+B)^2 - (A-B)^2 = 4AB \) is a powerful shortcut. Here, A is \( x \) and B is \( (y+z) \). If you don't spot the shortcut, a detailed expansion and cancellation of terms will also work, but it requires more steps and careful sign management.

Question 20. सिद्ध कीजिए कि \( (4x + 2y)^3 – (4x – 2y)^3 = 16y^3 + 192x^2y \)
Answer:
हलः
माना \( a = 4x + 2y \) और \( b = 4x – 2y \).
L.H.S. \( = a^3 - b^3 \)
हम जानते हैं कि \( a^3 - b^3 = (a - b)(a^2 + b^2 + ab) \)
पहले \( a-b \), \( a^2 \), \( b^2 \), और \( ab \) ज्ञात करें:
\( a - b = (4x + 2y) - (4x - 2y) = 4x + 2y - 4x + 2y = 4y \)
\( a^2 = (4x + 2y)^2 = (4x)^2 + (2y)^2 + 2(4x)(2y) = 16x^2 + 4y^2 + 16xy \)
\( b^2 = (4x - 2y)^2 = (4x)^2 + (2y)^2 - 2(4x)(2y) = 16x^2 + 4y^2 - 16xy \)
\( ab = (4x + 2y)(4x - 2y) = (4x)^2 - (2y)^2 = 16x^2 - 4y^2 \)
अब इन मानों को \( a^3 - b^3 \) के सूत्र में रखें:
\( a^3 - b^3 = (4y)[(16x^2 + 4y^2 + 16xy) + (16x^2 + 4y^2 - 16xy) + (16x^2 - 4y^2)] \)
\( = (4y)[16x^2 + 4y^2 + 16xy + 16x^2 + 4y^2 - 16xy + 16x^2 - 4y^2] \)
पदों को संयोजित करें:
\( = (4y)[(16x^2 + 16x^2 + 16x^2) + (4y^2 + 4y^2 - 4y^2) + (16xy - 16xy)] \)
\( = (4y)[48x^2 + 4y^2 + 0] \)
\( = 4y(48x^2 + 4y^2) \)
\( = 192x^2y + 16y^3 \)
\( = 16y^3 + 192x^2y = \text{R.H.S.} \)
In simple words: To prove the given identity, we use the difference of cubes formula: \( a^3-b^3 = (a-b)(a^2+b^2+ab) \). We define \( a=(4x+2y) \) and \( b=(4x-2y) \). We then calculate \( (a-b) \), \( a^2 \), \( b^2 \), and \( ab \) individually. Substituting these into the formula and simplifying the resulting expression proves the identity.

🎯 Exam Tip: This problem is simplified by using the identity \( a^3-b^3=(a-b)(a^2+b^2+ab) \). Carefully calculate each component (\( a-b \), \( a^2 \), \( b^2 \), \( ab \)) separately before substituting them into the main formula. Pay close attention to cancellations of terms like \( +16xy \) and \( -16xy \).

Question 21. सिद्ध कीजिए कि \( 7x^3 + 8y^3 -(4x + 3y)(16x^2 – 12xy + 9y^2) = -57x^3 – 19y^3 \)
Answer:
हलः
L.H.S. \( = 7x^3 + 8y^3 – (4x + 3y)(16x^2 – 12xy + 9y^2) \)
हम जानते हैं कि \( (a+b)(a^2-ab+b^2) = a^3+b^3 \).
यहाँ, \( a = 4x \) और \( b = 3y \).
तो, \( (4x + 3y)(16x^2 – 12xy + 9y^2) = (4x)^3 + (3y)^3 \)
\( = 64x^3 + 27y^3 \)
अब L.H.S. में इस मान को रखने पर:
\( = 7x^3 + 8y^3 – (64x^3 + 27y^3) \)
\( = 7x^3 + 8y^3 – 64x^3 - 27y^3 \)
\( = (7x^3 - 64x^3) + (8y^3 - 27y^3) \)
\( = -57x^3 - 19y^3 = \text{R.H.S.} \)
In simple words: The key to proving this identity is recognizing that \( (4x+3y)(16x^2-12xy+9y^2) \) is the expansion of \( (4x)^3+(3y)^3 \). Once this cubic sum is identified and substituted back into the original expression, it becomes a simple subtraction of like terms, leading to the right-hand side.

🎯 Exam Tip: Spotting the sum/difference of cubes pattern \( (a+b)(a^2-ab+b^2) = a^3+b^3 \) or \( (a-b)(a^2+ab+b^2) = a^3-b^3 \) is crucial here. Apply this identity first to simplify the bracketed term, then perform the final subtraction. Be careful with distributing the negative sign.

Balaji Publications Mathematics Class 9 Solutions