UP Board Solutions Class 9 Maths Chapter 4 Algebraic Identities Ex 4.3

Ex 4.3 Algebraic Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. (2x + + 1)3 का मान ज्ञात कीजिए।
Answer: हलः
\((2x + 1)^3 = (2x)^3 + (1)^3 + 3 \times 2x \times 1(2x + 1)\)
\(= 8x^3 + 1 + 12x^2 + 6x\)
In simple words: This problem involves expanding a binomial expression \((a+b)^3\) using the algebraic identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). We substitute \(a=2x\) and \(b=1\) into the formula and simplify.

🎯 Exam Tip: Remember to correctly apply the binomial cube identity and perform all multiplications accurately to avoid errors in coefficients and powers.

Question 2. (2a - 3b) का मान ज्ञात कीजिए।
Answer: हलः \((2a - 3b)^3 = (2a)^3 + (-3b)^3 + 3(2a) (-3b)(2a - 3b)\)
\(= 8a^3 - 27b^3 - 18ab(2a - 3b)\)
\(= 8a^3 - 27b^3 - 36a^2b + 54ab^2\)
In simple words: Here, we expand \((a-b)^3\) using the identity \((a-b)^3 = a^3 - b^3 - 3ab(a-b)\). Substitute \(a=2a\) and \(b=3b\) into the formula, carefully managing the negative signs.

🎯 Exam Tip: Pay close attention to the signs when working with negative terms in the expansion, especially when squaring or cubing them, and during multiplication.

Ex 4.3 Algebraic Identities लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

Question 3. निम्न व्यंजकों के घन का मान ज्ञात कीजिए ।
Answer: हलः
(i) \(\left(\frac{1}{x}+\frac{y}{3}\right)^3 = \left(\frac{1}{x}\right)^3 + \left(\frac{y}{3}\right)^3 + 3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)\left(\frac{1}{x}+\frac{y}{3}\right)\)
\(= \frac{1}{x^3} + \frac{y^3}{27} + \frac{y}{x^2} + \frac{y^2}{3x}\)
(ii) \(\left(4-\frac{1}{3x}\right)^3 = (4)^3 + \left(-\frac{1}{3x}\right)^3 + 3(4)\left(-\frac{1}{3x}\right)\left(4-\frac{1}{3x}\right)\)
\(= 64 - \frac{1}{27x^3} - \frac{16}{x} + \frac{4}{x^2}\)
(iii) \((3x-2y)^3 = (3x)^3 + (-2y)^3 + 3(3x)(-2y)(3x-2y)\)
\(= 27x^3 - 8y^3 - 18xy(3x-2y)\)
\(= 27x^3 - 8y^3 - 54x^2y + 36xy^2\)
In simple words: This question requires applying the binomial cube identities \((a+b)^3\) and \((a-b)^3\) to expressions involving fractions and multiple variables. Each part involves substituting the given terms into the appropriate identity and simplifying the resulting expression.

🎯 Exam Tip: When dealing with fractional terms, ensure correct exponentiation for both numerator and denominator. Be meticulous with algebraic simplification, especially when combining like terms.

Question 4. निम्न को सरल कीजिए।
Answer: हलः
(i) \((4x + 2y)^3 + (4x - 2y)^3\)
\((4x + 2y)^3 = (4x)^3 + (2y)^3 + 3 \cdot (4x) \cdot (2y) (4x + 2y)\)
\(= 64x^3 + 8y^3 + 24xy (4x + 2y)\)
\(= 64x^3 + 8y^3 + 96x^2y + 48xy^2\)
\((4x - 2y)^3 = (4x)^3 + (-2y)^3 + 3 \cdot (4x) (-2y) (4x - 2y)\)
\(= 64x^3 - 8y^3 - 24xy (4x - 2y)\)
\(= 64x^3 - 8y^3 - 96x^2y + 48xy^2\)
इसलिए \((4x + 2y)^3 + (4x - 2y)^3 = 128x^3 +96xy^2\)
(ii) \((4x + 2y)^3 - (4x - 2y)^3\)
\(= (64x^3 + 8y^3 + 96x^2y + 48xy^2) - (64x^3 - 8y^3 - 96x^2y + 48xy^2)\)
\(= 64x^3 + 8y^3 + 96x^2y + 48xy^2 - 64x^3 + 8y^3 + 96x^2y - 48xy^2\)
\(= 16y^3 + 192x^2y\)
(iii) \((a + 3)^3 + (a - 3)^3\)
\((a + 3)^3 = a^3 + 27 + 3a \cdot 3(a + 3)\)
\(= a^3 + 27 + 9a^2 + 27a\)
\((a - 3)^3 = a^3 - 27 - 3a \cdot 3(a - 3)\)
\(= a^3 - 27 - 9a^2 + 27a\)
\(\therefore (a + 3)^3 + (a - 3)^3 = 2a^3 +54a\)
In simple words: This question involves simplifying expressions that are sums or differences of two cubed binomials. We first expand each binomial individually using the \((a+b)^3\) and \((a-b)^3\) identities, and then combine the results according to the operation (addition or subtraction).

🎯 Exam Tip: When subtracting expanded expressions, be careful to distribute the negative sign to every term inside the parentheses. Look for terms that cancel each other out to simplify the final result.

Ex 4.3 Algebraic Identities लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

Question 5. यदि \(a - \frac{1}{a}\) = 7, तब \(a^3 - \frac{1}{a^3}\) का मान ज्ञात कीजिए।
Answer: हलः
\(a - \frac{1}{a} = 7\)
वर्ग करने पर
\(a^2 + \frac{1}{a^2} - 2a \cdot \frac{1}{a} = 49\)
\(a^2 + \frac{1}{a^2} = 49 + 2 = 51\)
\(a^3 - \frac{1}{a^3} = \left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}+1\right)\)
\(= (7)[51 + 1] = 7[52] = 364\)
In simple words: Given \(a - \frac{1}{a}\), we need to find \(a^3 - \frac{1}{a^3}\). First, square the given expression to find \(a^2 + \frac{1}{a^2}\). Then, use the factorization identity for difference of cubes, \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\), with \(b = \frac{1}{a}\) to find the required value.

🎯 Exam Tip: This problem demonstrates the interrelation of algebraic identities. Knowing \((x-y)^2\) and \((x^3-y^3)\) factorizations is crucial. Practice such chain calculations for speed and accuracy.

Question 6. यदि a - b = 4 व ab = 21, तब \(a^3 - b^3\) का मान ज्ञात कीजिए।
Answer: हलः
\(a - b = 4\)
वर्ग करने पर
\(a^2 + b^2 - 2ab = 16\)
\(a^2 + b^2 - 2 \times 21 = 16\)
\(a^2 + b^2 = 16 + 42 = 58\)
अब \(a^3 - b^3 = (a - b)(a^2 + b^2 + ab) = (4)(58 + 21) = (4)(79) = 316\)
In simple words: To find \(a^3 - b^3\) when \(a-b\) and \(ab\) are given, first use \((a-b)^2\) to find \(a^2+b^2\). Then, substitute the values of \(a-b\), \(a^2+b^2\), and \(ab\) into the factorization formula for the difference of cubes.

🎯 Exam Tip: This is a standard type of problem where you build up required components for a complex identity. Remember the steps: finding \(a^2+b^2\) first, then using the cubic factorization.

Question 7. यदि x + y = 10 व xy = 21, तब \(x^3 + y^3\) का मान ज्ञात कीजिए।
Answer: हलः
\(x + y = 10 \quad \ldots \ldots \ldots \ldots (1)\)
घन करने पर
\((x + y)^3 = (10)^3\)
\(x^3 + y^3 + 3xy (x + y) = 1000\)
\(x^3 + y^3 + 3 \cdot 21 (10) = 1000\)
\(x^3 + y^3 + 630 = 1000\)
\(x^3 + y^3 = 1000 - 630 = 370\)
In simple words: Given the sum and product of two variables, we need to find the sum of their cubes. We achieve this by cubing the sum \((x+y)\) and then substituting the known values of \(x+y\) and \(xy\) into the expanded form of the identity.

🎯 Exam Tip: This problem efficiently uses the identity \((x+y)^3 = x^3+y^3+3xy(x+y)\). Recognising this direct application saves time compared to finding x and y individually.

Question 8. यदि 3x - 2y =11 व xy = 12, तब \(27x^3 - 8y^3\) का मान ज्ञात कीजिए।
Answer: हलः
\(\therefore 3x - 2y = 11 \quad \ldots \ldots \ldots \ldots (1)\)
वर्ग करने पर
\(9x^2 + 4y^2 - 12xy = 121\)
\(9x^2 + 4y^2 - 12 \times 12 = 121\)
\(9x^2 + 4y^2 = 121 + 144 = 265\)
\(27x^3 - 8y^3 = (3x)^3 - (2y)^3\)
\(= (3x - 2y)(9x^2 + 4y^2 + 6xy) = (11)(265 + 6 \times 12) = (11)(265 + 72) = (11) (337) = 3707\)
In simple words: To find \(27x^3 - 8y^3\), which is \((3x)^3 - (2y)^3\), we first find \((3x-2y)\) and \((9x^2+4y^2)\). Squaring the given equation \((3x-2y)=11\) helps find \(9x^2+4y^2\), then we use the difference of cubes formula.

🎯 Exam Tip: This problem effectively uses both squaring a binomial and the difference of cubes identity. It's essential to recognize \(27x^3\) as \((3x)^3\) and \(8y^3\) as \((2y)^3\) for correct application.

Question 9. यदि \(a^2 + \frac{1}{a^2}\) = 98, तब \(a^3 + \frac{1}{a^3}\) का मान ज्ञात कीजिए।
Answer: हलः
\(\left(a+\frac{1}{a}\right)^2= a^2+\frac{1}{a^2}+2\)
\(\left(a+\frac{1}{a}\right)^2 = 98 + 2 = 100\)
\(a + \frac{1}{a} = \sqrt{100} = 10\)
\(a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right)\)
\(= (10) (98-1) = 970\)
In simple words: Given \(a^2 + \frac{1}{a^2}\), we need to find \(a^3 + \frac{1}{a^3}\). First, find \(a + \frac{1}{a}\) using the identity \((a+\frac{1}{a})^2 = a^2+\frac{1}{a^2}+2\). Then, apply the sum of cubes factorization \(x^3+y^3 = (x+y)(x^2-xy+y^2)\) with \(x=a\) and \(y=\frac{1}{a}\).

🎯 Exam Tip: This problem is a multi-step application of identities. Accurately deriving \(a+\frac{1}{a}\) from \(a^2+\frac{1}{a^2}\) is the first critical step before applying the sum of cubes identity.

Question 10. यदि \(a^2 + \frac{1}{a^2}\) = 51, तब \(a^3 - \frac{1}{a^3}\) का मान ज्ञात कीजिए।
Answer: हल:
\(\left(a-\frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2\)
\(\left(a-\frac{1}{a}\right)^2 = 51 - 2 = 49\)
\(a-\frac{1}{a} = \sqrt{49} = 7\)
\(a^3-\frac{1}{a^3}=\left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}+1\right)\)
\(= (7)(51+1) = (7)(52) = 364\)
In simple words: To find \(a^3 - \frac{1}{a^3}\) from \(a^2 + \frac{1}{a^2}\), first calculate \(a - \frac{1}{a}\) using the identity \((a-\frac{1}{a})^2 = a^2+\frac{1}{a^2}-2\). Then, use the difference of cubes formula, \(x^3-y^3 = (x-y)(x^2+xy+y^2)\), with \(x=a\) and \(y=\frac{1}{a}\).

🎯 Exam Tip: Be careful with the signs when moving from \(a^2+\frac{1}{a^2}\) to \(a-\frac{1}{a}\) versus \(a+\frac{1}{a}\). The difference of cubes identity requires \(a-\frac{1}{a}\).

Ex 4.3 Algebraic Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

Question 11. निम्न के मान ज्ञात कीजिए ।
Answer: हलः
(i) \((99)^3 = (100 – 1)^3\)
\(= (100)^3 - (1)^3 – 3 \times 100 \times 1(100 – 1) = 1000000 – 1 – 30000 + 300 = 970299\)
(ii) \((9.9)^3 = (10 – 0.1)^3\)
\(= (10)^3 – (0.1)^3 – 3 \times 10 \times 0.1(10 – 0.1) = 1000 – 0.001 – 3(9.9) = 1000 – 0.001 – 29.7 = 970.299\)
(iii) \((10.4)^3 = (10 + 0.4)^3\)
\(= (10)^3 + (0.4)^3 + 3 \times 10 \times 0.4(10 + 0.4) = 1000 + 0.064 + 12(10.4) = 1000 + 0.064 + 124.8 =1124.864\)
(iv) \((598)^3 = (600-2)^3\)
\(= (600)^3 – (2)^3 – 3 \times 600 \times 2(600 – 2) = 216000000 – 8-21,60,000 + 7200 =21,38,47,192\)
(v) \((402)^3 = (400 + 2)^3\)
\(= (400)^3 + (2)^3 + 3 \times 400 \times 2(400 + 2) = 64000000 + 8 + 960000 + 4800 = 64964808\)
(vi) \((1002)^3 = (1000 + 2)^3\)
\(= (1000)^3 + (2)^3 + 3 \times 1000 \times 2(1000 + 2) = 1000000000 + 8+ 6000000 + 12000 = 1006012008\)
In simple words: This problem involves calculating the cube of numbers that are close to multiples of 10, 100, or 1000. By expressing them as \((x \pm y)^3\), we can use algebraic identities \((a+b)^3\) or \((a-b)^3\) to simplify calculations, avoiding direct multiplication of large numbers.

🎯 Exam Tip: This technique is a crucial shortcut for mental math and exam efficiency. Always look for ways to express numbers as sums or differences of easier-to-cube values (like 10, 100, 1000) to apply identities.

Question 12. निम्न के मान ज्ञात कीजिए।
Answer: हलः
(i) \((46)^3 + (34)^3\)
\(= (40 + 6)^3 + (40-6)^3\)
\(= (40)^3 + (6)^3 + 3 \times 40 \times 6(40 + 6) + (40)^3 - (6)^3 - 3 \times 40 \times 6(40 - 6)\)
\(= 2(40)^3 + 3 \times 40 \times 6[46 - 34]\)
\(= 2(40)^3 + 720 \times 12 = 128000 + 8640 = 136640\)
(ii) \((23)^3 - (17)^3\)
\(= (20 + 3)^3 - (20 - 3)^3\)
\(= (20)^3 + (3)^3 + 3 \times 20 \times 3(20 + 3) - [(20)^3 + (-3)^3 + 3 \times 20 \times (-3)(20 - 3)]\)
\(= (20)^3 + (3)^3 + 3 \times 20 \times 3(20 + 3) - (20)^3 + (3)^3 + 3 \times 20 \times 3(20 - 3)\)
\(= 2(3)^3 + 3 \times 20 \times 3(23 + 17) = 2 \times 27 + 180(40) = 54+ 7200 = 7254\)
(iii) \((111)^3 - (89)^3\)
\(= (100 + 11)^3 - (100 - 11)^3\)
\(= (100)^3 + (11)^3 + 3 \times 100 \times 11 (100 + 11) - [(100)^3 + (-11)^3 + 3 \times 100 \times (-11)(100 - 11)]\)
\(= (100)^3 + (11)^3 + 3 \times 100 \times 11(100 + 11) - (100)^3 + (11)^3 + 3 \times 100 \times 11(100 - 11)\)
\(= 2(11)^3 + 3 \times 100 \times 11(111 + 89) = 2(1331) + 3 \times 100 \times 11 \times 200 = 2662 + 660000 = 662662\)
In simple words: This problem asks us to evaluate expressions of the form \((a+b)^3 \pm (a-b)^3\). By rewriting the numbers in terms of a central value (like 40, 20, 100) plus or minus an offset, we can use the identities \((X+Y)^3\) and \((X-Y)^3\) and then simplify the sum or difference.

🎯 Exam Tip: Recognizing patterns like \(A^3 + B^3\) where \(A=X+Y\) and \(B=X-Y\) (or similar for subtraction) allows for strategic use of identities. This simplifies calculations significantly by cancelling many terms.

Question 13. यदि 3x + 2y = 20 व \(xy = \frac{14}{9}\) तब \(27x^3 + 8y^3\) का मान ज्ञात कीजिए।
Answer: हलः
\(3x + 2y = 20 \quad \ldots \ldots \ldots \ldots (1)\)
वर्ग करने पर
\(9x^2 + 4y^2 + 12xy = 400\)
\(9x^2 +4y^2 = 400 - 12xy\)
\(27x^3 + 8y^3\)
\(= (3x)^3 + (2y)^3\)
\(= (3x + 2y) (9x^2 + 4y^2 - 6xy) = (3x + 2y)(400 - 12xy - 6xy) = (3x + 2y)(400 - 18xy) = (20)(400 - 18 \times \frac{14}{9}) = (20) (400 - 28) = (20) (372) = 7440\)
In simple words: To find \(27x^3 + 8y^3\), which is \((3x)^3 + (2y)^3\), given \(3x+2y\) and \(xy\), we first find \(9x^2+4y^2\) by squaring \((3x+2y)\). Then, substitute these values into the sum of cubes factorization formula: \(a^3+b^3 = (a+b)(a^2-ab+b^2)\).

🎯 Exam Tip: This problem involves a sequence of identity applications. Accurately deriving intermediate values like \(9x^2+4y^2\) from the given information is key to successfully applying the sum of cubes identity.

Question 14.
Answer: हलः
\(m^4 + \frac{1}{m^4} = 194\)
\(\left(m^2+\frac{1}{m^2}\right)^2 = m^4 + \frac{1}{m^4} + 2\)
\(\left(m^2 + \frac{1}{m^2}\right)^2 = 194 + 2 = 196\)
\(m^2 + \frac{1}{m^2} = \sqrt{196} = 14\)
\(\left(m + \frac{1}{m}\right)^2 = m^2 + \frac{1}{m^2} + 2\)
\(\left(m + \frac{1}{m}\right)^2 = 14 + 2 = 16\)
\(m + \frac{1}{m} = \sqrt{16} = 4\)
इसी प्रकार
\(\left(m+\frac{1}{m}\right)^3 = m^3 + \frac{1}{m^3} + 3m\left(\frac{1}{m}\right)\left(m+\frac{1}{m}\right)\)
\((4)^3 = m^3 + \frac{1}{m^3} + 3 \times 4\)
\(64 - 12 = m^3 + \frac{1}{m^3} \quad \text{या} \quad m^3 + \frac{1}{m^3} = 52\)
In simple words: Given \(m^4 + \frac{1}{m^4}\), we need to find \(m^3 + \frac{1}{m^3}\). This is a multi-step process. First, use identities to find \(m^2 + \frac{1}{m^2}\), then \(m + \frac{1}{m}\). Finally, cube \(m + \frac{1}{m}\) to derive the value of \(m^3 + \frac{1}{m^3}\).

🎯 Exam Tip: This problem tests your ability to work backward and forward through algebraic identities. Remember the sequential pattern: from \(a^4+\frac{1}{a^4}\) to \(a^2+\frac{1}{a^2}\) to \(a+\frac{1}{a}\) and then to \(a^3+\frac{1}{a^3}\).

Question 15.
Answer: हलः
\(a+\frac{1}{a}=3\)
वर्ग करने पर
\(a^2+\frac{1}{a^2}+2=9\)
\(a^2+\frac{1}{a^2}=9-2=7\)
घन करने पर
\(\left(a+\frac{1}{a}\right)^3=(3)^3\)
\(a^3+\frac{1}{a^3}+3a\left(\frac{1}{a}\right)\left(a+\frac{1}{a}\right) = 27\)
\(a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right) = 27\)
\(a^3+\frac{1}{a^3}+3\times 3 = 27\)
\(a^3+\frac{1}{a^3}=27-9=18\)
\(a^2+\frac{1}{a^2}=7\)
In simple words: This problem first requires finding \(a^2+\frac{1}{a^2}\) by squaring the given expression \(a+\frac{1}{a}\). Then, to find \(a^3+\frac{1}{a^3}\), cube the expression \(a+\frac{1}{a}\) and substitute its value into the expanded identity.

🎯 Exam Tip: This is a fundamental problem combining squaring and cubing binomials. Practice these steps until they are intuitive, as they form the basis for more complex problems.

Question 16.
Answer: हलः
\(y-\frac{1}{y} = 3+ 2\sqrt{2}\)
वर्ग करने पर
\(y^2+\frac{1}{y^2}-2y \times \frac{1}{y} = (3+2\sqrt{2})^2\)
\(y^2+\frac{1}{y^2}-2 = 9+8+12\sqrt{2}\)
\(y^2+\frac{1}{y^2} = 17+12\sqrt{2}+2=19+12\sqrt{2}\)
\(y^3-\frac{1}{y^3}=\left(y-\frac{1}{y}\right)\left(y^2+\frac{1}{y^2}+1\right)\)
\(= (3 + 2\sqrt{2}) (19+12\sqrt{2}+1)\)
\(= (3 + 2\sqrt{2})(20+12\sqrt{2})\)
\(= 60 + 36\sqrt{2} + 40\sqrt{2} + 48 = 108+76\sqrt{2}\)
In simple words: Given \(y-\frac{1}{y}\) with a surd expression, we first find \(y^2+\frac{1}{y^2}\) by squaring the given expression. Remember to expand \((a+b)^2\) for the surd term. Then, use the difference of cubes identity \(a^3-b^3=(a-b)(a^2+ab+b^2)\) to find \(y^3-\frac{1}{y^3}\).

🎯 Exam Tip: When squaring expressions with surds, use the \((a+b)^2 = a^2+2ab+b^2\) identity carefully. Ensure accurate multiplication and combination of surd terms in the final step.