UP Board Solutions Class 9 Maths Chapter 5 Polynomial and their Factors Ex 5.7

Balaji Class 9 Maths Solutions Chapter 5 Polynomial And Their Factors Ex 5.7 बहुपद तथा उनके गुणनखण्ड

Question 1. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) x3 - 8y3
(ii) a3 - 0.216
(iii) 16a4 + 54a
(iv) a6 - 7a3 - 8
Answer:
(i) \( x^3 - 8y^3 = (x)^3 - (2y)^3 = (x - 2y)(x^2 + 4y^2 + 2xy) \)
(ii) \( a^3 - 0.216 = (a)^3 - (0.6)^3 = (a - 0.6)(a^2 - 0.36 + 0.6 a) \)
(iii) \( 16a^4 + 54a = 2a(8a^3 + 27) = 2a[(2a)^3 + (3)^3] = 2a[(2a + 3)(4a^2 + 9 - 6a)] \)
(iv) \( a^6 - 7a^3 - 8 = a^6 - 8a^3 + a^3 - 8 \)
\( = a^3 (a^3 - 8) + 1 (a^3 - 8) \)
\( = (a^3 - 8)(a^3 + 1) \)
\( = [(a)^3 - (2)^3][(a)^3 + (1)^3] \)
\( = (a - 2)(a^2 + 4 + 2a)(a + 1)(a^2 - a + 1) \)
In simple words: This question requires factoring various algebraic expressions using identities like \( a^3 - b^3 \) and by grouping terms. Recognize common factors and apply the appropriate factorization formulas.

🎯 Exam Tip: Mastering algebraic identities such as \( a^3 \pm b^3 \) is crucial for quickly solving factorization problems. Always look for common factors first before applying identities.

Question 2. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) x3 + 64
(ii) 1 - 125y3
(iii) \( \frac{p^{3}}{343} \) + 8q3
(iv) 7m4n - 7mn4
(v) x7y7 - xy
Answer:
(i) \( x^3 + 64 = (x)^3 + (4)^3 = (x + 4)(x^2 + 16 - 4x) \)
(ii) \( 1 - 125y^3 = (1)^3 - (5y)^3 = (1 - 5y)(1 + 25y^2 + 5y) \)
(iii) \( \frac{p^{3}}{343}+8 q^{3}=\left(\frac{p}{7}\right)^{3}+(2 q)^{3}=\left(\frac{p}{7}+2 q\right)\left(\frac{p^{2}}{49}+4 q^{2}-\frac{2}{7} p q\right) \)
(iv) \( 7m^4n - 7mn^4 = 7mn(m^3 - n^3) = 7mn[(m)^3 - (n)^3] = 7mn[(m - n)(m^2 + n^2 + mn)] \)
(v) \( x^7y^7 - xy = xy(x^6y^6 - 1) = xy[(x^2y^2)^3 - (1)^3] \)
\( = xy[(x^2y^2 - 1)(x^4y^4 + 1 + x^2y^2)] \)
\( = xy [(xy)^2 - (1)^2] (x^4y^4 - 1 + x^2y^2) \)
\( = xy (xy + 1)(xy - 1)[(x^2y^2 + 1)^2 - (xy)^2] \)
\( = xy(xy + 1)(xy - 1)(x^2y^2 + 1 + xy)(x^2y^2 + 1 - xy) \)
In simple words: This question involves factoring various polynomial expressions using standard algebraic identities like sum/difference of cubes and by identifying common factors. Some problems require multiple steps of factorization.

🎯 Exam Tip: When dealing with higher powers, try to express them as cubes or squares of simpler terms to apply standard identities. For expressions with multiple terms, look for common factors first to simplify the problem.

Question 3. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) (a + 2b)³ - (a - b)3
(ii) a3 + b3 + c(a² - ab + b²)
(iii) x6 - 7x3 - 8
(iv) x3 - 3x2 + 3x + 7
Answer:
(i) \( (a + 2b)^3 - (a - 2b)^3 = (a + 2b - (a - 2b)) [(a + 2b)^2 + (a - 2b)^2 + (a + 2b) (a - 2b)] \)
\( = (4b)[a^2 + 4b^2 + 4ab + a^2 + 4b^2 - 4ab + a^2 - 4b^2] \)
\( = (4b)[3a^2 + 4b^2] \)
(ii) \( a^3 + b^3 + c(a^2 - ab + b^2) = (a + b)(a^2 - ab + b^2) + c(a^2 - ab + b^2) \)
\( = (a^2 - ab + b^2)[a + b + c] \)
(iii) \( x^6 - 7x^3 - 8 = x^6 - (8 - 1)x^3 - 8 \) ( \( 8 = 1 \times 8 \) )
\( = x^6 - 8x^3 + x^3 - 8 = x^3 (x^3 - 8) + 1(x^3 - 8) = (x^3 - 8)(x^3 + 1) \)
\( = [(x)^3 - (2)^3][(x)^3 + (1)^3] = (x - 2)(x^2 + 4 + 2x)(x + 1)(x^2 + 1 - x) \)
(iv) \( x^3 - 3x^2 + 3x + 7 \)
माना \( f(x) = x^3 - 3x^2 + 3x + 7 \)
\( x + 1 = 0 \)
\( x = -1 \) रखने पर,
\( = (-1)^3 - 3(-1)^2 + 3 \times (-1) +7 \)
\( = -1 - 3 - 3 + 7 = -7 + 7 = 0 \)
\( \therefore (x + 1) \), \( f(x) \) का एक गुणनखण्ड है।
अतः
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बहुपद के भाग का चित्रण है, जहाँ \( x^3 - 3x^2 + 3x + 7 \) को \( x+1 \) से विभाजित किया गया है। भागफल \( x^2 - 4x + 7 \) है और शेषफल शून्य है।
अतः \( x^3 - 3x^2 + 3x + 7 \) के गुणनखण्ड \( (x + 1)(x^2 - 4x + 7) \) है।
In simple words: This question focuses on factoring various polynomials, including those involving the sum/difference of cubes and polynomial long division. For expressions with \( x^3 - 3x^2 + 3x + 7 \), synthetic division or trial and error to find a root (like \( x = -1 \)) helps identify one factor, then long division provides the rest.

🎯 Exam Tip: For polynomial factorization, remember the Factor Theorem: if \( f(a) = 0 \), then \( (x - a) \) is a factor. For cubic expressions like \( a^3 \pm b^3 \), applying the correct identity directly saves time. Polynomial long division is a key skill for dividing by known factors.

Question 4. निम्न व्यंजकों को सरल करके उनके मान ज्ञात कीजिए
(i) \( \frac{0.96 \times 0.96 \times 0.96 + 0.04 \times 0.04 \times 0.04}{0.96 \times 0.96 - 0.96 \times 0.04 + 0.04 \times 0.04} \)
(ii) \( \frac{173 \times 173 \times 173 + 127 \times 127 \times 127}{173 \times 173 - 173 \times 127 + 127 \times 127} \)
Answer:
(i) \( \frac{0.96 \times 0.96 \times 0.96 + 0.04 \times 0.04 \times 0.04}{0.96 \times 0.96 - 0.96 \times 0.04 + 0.04 \times 0.04} \)
माना \( 0.96 = a \) तथा \( 0.04 = b \)
\( = \frac{a \times a \times a + b \times b \times b}{a \times a - a \times b + b \times b} \)
\( = \frac{a^3 + b^3}{a^2 - ab + b^2} \)
\( = \frac{(a + b)(a^2 + b^2 - ab)}{(a^2 - ab + b^2)} \)
\( = a + b = 0.96 + 0.04 = 1.00 \)
(ii) \( \frac{173 \times 173 \times 173 + 127 \times 127 \times 127}{173 \times 173 - 173 \times 127 + 127 \times 127} \)
माना \( a = 173 \) तथा \( b = 127 \)
\( = \frac{a \times a \times a + b \times b \times b}{a \times a - a \times b + b \times b} \)
\( = \frac{a^3 + b^3}{a^2 - ab + b^2} \)
\( = \frac{(a + b)(a^2 + b^2 - ab)}{(a^2 - ab + b^2)} \)
\( = a + b = 173 + 127 = 300 \)
In simple words: This question demonstrates how to simplify complex numerical fractions by recognizing the algebraic identity \( \frac{a^3 + b^3}{a^2 - ab + b^2} = a + b \). By substituting the given numbers with variables, the expressions simplify to a straightforward sum.

🎯 Exam Tip: Always look for underlying algebraic identities when simplifying numerical expressions. Recognizing patterns like \( a^3 + b^3 \) in complex fractions can significantly reduce calculation time and prevent errors.

Ex 5.7 Polynomial And Their Factors स्वमूल्यांकन परीक्षण (Self Assessment Test)

Question 1. निम्न के गुणनखण्ड कीजिए। \( 27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p \)
Answer: \( (3 p)^{3}+\left(-\frac{1}{6}\right)^{3}+3(3 p)\left(-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)=\left(3 p-\frac{1}{6}\right)^{3} \)
In simple words: This problem involves factoring a polynomial expression by recognizing the identity for a perfect cube: \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \). Identify \( a \) and \( b \) to directly apply the formula.

🎯 Exam Tip: When terms resemble parts of a cubic expansion, try to fit them into the \( (a \pm b)^3 \) identity. Practice identifying \( a \) and \( b \) from expressions like \( 27p^3 \) and \( -\frac{1}{216} \).

Question 2. 49a2 + 70ab + 25b2
Answer:
\( (7a)^2 + 2 \times 7a \times 5b + (5b)^2 \)
सूत्र \( x^2 + 2xy + y^2 = (x + y)^2 \) से
\( = (7a + 5b)^2 \)
In simple words: This expression is a perfect square trinomial. By recognizing the pattern \( a^2 + 2ab + b^2 \), it can be factored directly into \( (a+b)^2 \).

🎯 Exam Tip: Look for three terms where the first and third are perfect squares and the middle term is twice the product of their square roots. This indicates a perfect square trinomial.

Question 3. 4x2 + y2 + z2 - 4xy -2yz + 4xz
Answer:
\( (2x)^2 + (-y)^2 + (z)^2 + 2 \cdot (2x)(-y) + 2(-y)(z) + 2 \cdot (2x)(z) \)
\( = (2x - y + z)^2 = (2x - y + z)(2x - y + z) \)
In simple words: This expression is a perfect square trinomial involving three variables. Recognize the pattern \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \) and identify the terms \( a, b, c \) (paying attention to signs) to factor it.

🎯 Exam Tip: For expressions with three squared terms and three double-product terms, consider the \( (a+b+c)^2 \) identity. The signs of the double-product terms will help determine the signs of \( a, b, \) and \( c \).

Question 4. 64m³ - 343n3
Answer:
\( 64m^3 - 343n^3 = (4m)^3 - (7n)^3 = (4m - 7n)(16m^2 + 49n^2 + 28mn) \)
In simple words: This problem is a direct application of the difference of cubes formula: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Identify \( a \) as \( 4m \) and \( b \) as \( 7n \).

🎯 Exam Tip: Memorize the difference of cubes formula. Identify \( a^3 \) and \( b^3 \) correctly, then substitute into the formula. Remember the positive sign in the \( ab \) term for difference of cubes.

Question 5. 27 - 125a3 - 135a + 225a²
Answer:
\( 27 - 125a^3 - 135a + 225a^2 \)
\( = (3)^3 - (5a)^3 - 45a(3 - 5a) \)
\( = (3 - 5a)(9 + 25a^2 + 15a) - 45a(3 - 5a) \)
\( = (3 - 5a)(9 + 25a^2 + 15a - 45a) \)
\( = (3 - 5a)(9 + 25a^2 - 30a) \)
\( = (3 - 5a)[(3)^2 + (5a)^2 - 2 \times 3 \times 5a] \)
\( = (3 - 5a)(3 - 5a)^2 = (3 - 5a)^3 \)
In simple words: This expression can be factored by recognizing it as a perfect cube expansion in the form of \( (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 \) or \( a^3 - b^3 - 3ab(a-b) \). Identify \( a = 3 \) and \( b = 5a \).

🎯 Exam Tip: When an expression has four terms that resemble a cubic expansion, try to rearrange and group them to fit the \( (a \pm b)^3 \) identity. Clearly identify \( a \) and \( b \) to apply the formula correctly.

Question 6. 8a3 - b3 - 12a2b + 6ab2
Answer:
\( 8a^3 - b^3 - 12a^2b + 6ab^2 \)
\( = (2a)^3 - (b)^3 - 6ab (2a - b) \)
\( = (2a - b)(4a^2 + b^2 + 2ab) - 6ab(2a - b) \)
\( = (2a - b) [4a^2 + b^2 + 2ab - 6ab] \)
\( = (2a - b)[4a^2 + b^2 - 4ab] \)
\( = (2a - b)[(2a)^2 + (b)^2 - 2 \times 2a \times b] \)
\( = (2a - b)(2a - b)^2 = (2a - b)^3 \)
In simple words: This polynomial is a perfect cube. By recognizing \( a = 2a \) and \( b = b \), the expression matches the expansion of \( (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 \).

🎯 Exam Tip: Look for the two cubed terms and the two terms involving \( a^2b \) and \( ab^2 \). This often signals a perfect cube. Grouping terms and applying the difference of cubes identity initially can also lead to the solution.

Question 7. 64a3 - 27b3 - 144a2b + 108ab2
Answer:
\( (4a)^3 - (3b)^3 - 36ab(4a - 3b) \)
\( = (4a - 3b)(16a^2 + 9b^2 + 12ab) - 36ab(4a - 3b) \)
\( = (4a - 3b)(16a^2 + 9b^2 + 12ab - 36ab) \)
\( = (4a - 3b)(16a^2 + 9b^2 - 24ab) \)
\( = (4a - 3b)[(4a)^2 + (3b)^2 - 2 \times 4a \times 3b] \)
\( = (4a - 3b)[4a - 3b]^2 = (4a - 3b)^3 \)
In simple words: This problem involves factoring a four-term polynomial that is a perfect cube. Identify \( a = 4a \) and \( b = 3b \), then apply the identity \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \).

🎯 Exam Tip: When dealing with such polynomial forms, quickly identify the cubic terms to determine \( a \) and \( b \). Then, check if the remaining terms fit the \( -3a^2b + 3ab^2 \) part of the expansion.

Question 8. 8a3 + b3 + 12a2b + 6ab2
Answer:
\( (2a)^3 + (b)^3 + 6ab(2a + b) \)
\( = (2a + b)(4a^2 + b^2 - 2ab) + 6ab(2a + b) \)
\( = (2a + b)[4a^2 + b^2 - 2ab + 6ab] \)
\( = (2a + b)[4a^2 + b^2 + 4ab] \)
\( = (2a + b)[(2a)^2 + (b)^2 + 2 \times 2a \times b] \)
\( = (2a + b)(2a + b)^2 = (2a + b)^3 \)
In simple words: This polynomial is a perfect cube. By recognizing \( a = 2a \) and \( b = b \), the expression matches the expansion of \( (a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 \).

🎯 Exam Tip: For expressions appearing as sum of cubes plus two other terms, check if they fit the \( (a+b)^3 \) formula. Factoring out common factors like \( 3ab \) can help simplify and reveal the perfect cube structure.

Question 9. \( 27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p \)
Answer: \( (3 p)^{3}+\left(-\frac{1}{6}\right)^{3}+3(3 p)\left(-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)=\left(3 p-\frac{1}{6}\right)^{3} \)
In simple words: This problem involves factoring a polynomial expression by recognizing the identity for a perfect cube: \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \). Identify \( a \) and \( b \) to directly apply the formula.

🎯 Exam Tip: When terms resemble parts of a cubic expansion, try to fit them into the \( (a \pm b)^3 \) identity. Practice identifying \( a \) and \( b \) from expressions like \( 27p^3 \) and \( -\frac{1}{216} \).

Question 10. 27x3 + y <3 + y³ + z3 - 9xyz
Answer:
\( (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z) \)
\( = (3x + y + z)[(3x)^2 + y^2 + z^2 - 3xy - yz - 3xz] \)
\( = (3x + y + z)[9x^2 + y^2 + z^2 - 3xy - yz - 3xz] \)
In simple words: This expression is a special type of factorization for the sum of three cubes minus three times their product, \( a^3 + b^3 + c^3 - 3abc \). It factors into \( (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \).

🎯 Exam Tip: Recognize the \( a^3 + b^3 + c^3 - 3abc \) pattern. Correctly identify \( a, b, \) and \( c \) from the given terms, then substitute them into the factorization formula directly.

Question 11. सिद्ध कीजिए कि a² + b² + 2(ab + bc + ca) = (a + b)(a + b + 2c)
Answer:
L.H.S. \( = a^2 + b^2 + 2(ab + bc + ca) \)
\( = a^2 + b^2 + 2ab + 2bc + 2ca \)
\( = (a + b)^2 + 2c(a + b) \)
\( = (a + b)[a + b + 2c] \)
\( = \) R.H.S.
In simple words: This proof involves expanding the left-hand side and grouping terms to reveal the common factor \( (a+b) \), which then simplifies to the right-hand side. The key is recognizing \( a^2+b^2+2ab \) as \( (a+b)^2 \).

🎯 Exam Tip: For "prove that" questions, start from one side (usually the more complex one) and manipulate it algebraically using identities until it matches the other side. Look for common algebraic patterns like perfect squares.

Question 12. सिद्ध कीजिए कि 8a3 - b3 - 4ax + 2bx = (2a - b)(4a2 + 2ab + b² - 2x)
Answer:
L.H.S. \( = 8a^3 - b^3 - 4ax + 2bx \)
\( = (2a)^3 - (b)^3 - 2x(2a - b) \)
\( = (2a - b)(4a^2 + b^2 + 2ab) - 2x(2a - b) \)
\( = (2a - b)(4a^2 + b^2 + 2ab - 2x) \)
\( = \) R.H.S.
In simple words: To prove this identity, start from the left-hand side and factor the difference of cubes \( 8a^3 - b^3 \). Then, factor out \( -2x \) from the remaining terms. This will reveal a common factor of \( (2a-b) \), leading to the right-hand side.

🎯 Exam Tip: When factoring complex expressions for proofs, identify common algebraic identities first (like difference of cubes). Then, look for common factors among the remaining terms to group them effectively.

Question 13. सिद्ध कीजिए कि \( \frac{64}{125} x^{3}-8-\frac{96}{25} x^{2}+\frac{48}{5} x=\left(\frac{4 x}{5}-2\right)^{3} \)
Answer:
L.H.S. \( = \frac{64}{125} x^3 - 8 - \frac{96}{25} x^2 + \frac{48}{5} x \)
\( = \left(\frac{4x}{5}\right)^3 - (2)^3 - 3\left(\frac{4x}{5}\right)^2(2) + 3\left(\frac{4x}{5}\right)(2)^2 \)
\( = \left(\frac{4x}{5}\right)^3 - (2)^3 - 3\left(\frac{16x^2}{25}\right)(2) + 3\left(\frac{4x}{5}\right)(4) \)
\( = \left(\frac{4x}{5}\right)^3 - (2)^3 - \frac{96x^2}{25} + \frac{48x}{5} \)
\( = \left(\frac{4x}{5}-2\right)^3 \)
\( = \) R.H.S.
In simple words: This proof requires recognizing the given polynomial as the expansion of a perfect cube, specifically \( (a-b)^3 \). By identifying \( a = \frac{4x}{5} \) and \( b = 2 \), you can confirm that the left-hand side matches the expansion \( a^3 - b^3 - 3a^2b + 3ab^2 \).

🎯 Exam Tip: When proving identities involving cubic terms and fractions, express all terms in a form that clearly shows \( a^3, b^3, -3a^2b, \) and \( +3ab^2 \). This makes it easier to match the expansion of \( (a-b)^3 \).

Question 14. सिद्ध कीजिए कि a3x3 - 3a2bx2 + 3ab2x - b³ = (ax - b)³
Answer:
L.H.S. \( = a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3 \)
\( = (ax)^3 - (b)^3 - 3abx(ax - b) \)
\( = (ax - b)(a^2x^2 + b^2 + abx) - 3abx(ax - b) \)
\( = (ax - b)(a^2x^2 + b^2 + abx - 3abx) \)
\( = (ax - b)(a^2x^2 + b^2 - 2abx) \)
\( = (ax - b)(ax - b)^2 = (ax - b)^3 \)
\( = \) R. H.S.
In simple words: This proof involves recognizing the left-hand side as the expansion of \( (a-b)^3 \). Identify \( a = ax \) and \( b = b \). The expression matches \( (ax)^3 - (b)^3 - 3(ax)^2(b) + 3(ax)(b)^2 \).

🎯 Exam Tip: A four-term polynomial with two perfect cube terms (like \( (ax)^3 \) and \( (b)^3 \)) and two intermediate terms often represents a perfect cube expansion. Verify if the intermediate terms match \( \pm 3a^2b \) and \( \pm 3ab^2 \).

Question 15. सिद्ध कीजिए कि a3 + b3 b3 + 1 + 3ab = (a - b + 1)(a3 + b³ + ab - a + b + 1)
Answer:
L.H.S. \( = (a)^3 + (-b)^3 + (1)^3 - 3(a)(-b)(1) \)
\( = (a - b + 1)(a^2 + (-b)^2 + 1^2 - (a)(-b) - (-b)(1) - (1)(a)) \)
\( = (a - b + 1)(a^2 + b^2 + 1 + ab + b - a) \)
[ \( \therefore x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \) ]
\( \therefore a^3 - b^3 + 1 + 3ab = (a - b + 1)(a^2 + b^2 + 1 + ab + b - a) \)
\( = \) R.H.S.
In simple words: This proof utilizes the sum of cubes identity \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \). By setting \( x=a, y=-b, \) and \( z=1 \), the left-hand side directly transforms into the right-hand side.

🎯 Exam Tip: The identity for \( a^3+b^3+c^3-3abc \) is powerful. When you see three cubed terms and a term involving \( 3 \times \) product of bases, consider this identity, paying close attention to signs for \( a,b,c \).

Question 16. यदि p = 2 - a, तब सिद्ध कीजिए कि a³ + 6ap + p3 - 8 = 0
Answer:
L.H.S. \( = a^3 + 6ap + p^3 - 8 \)
\( = a^3 + p^3 - 8 + 6ap \)
From \( p = 2 - a \)
\( \implies p + a = 2 \)
We know that if \( a+b+c = 0 \), then \( a^3+b^3+c^3 = 3abc \).
Here, let \( x = a, y = p, z = -2 \).
Then \( x+y+z = a+p-2 \). Since \( a+p=2 \), \( x+y+z = 2-2 = 0 \).
So, \( a^3 + p^3 + (-2)^3 = 3(a)(p)(-2) \)
\( a^3 + p^3 - 8 = -6ap \)
\( a^3 + p^3 - 8 + 6ap = 0 \)
Alternatively, substituting \( p = 2 - a \) into LHS:
\( = a^3 + (2 - a)^3 - 8 + 6a(2 - a) \)
\( = a^3 + (2^3 - a^3 - 3 \cdot 2^2 \cdot a + 3 \cdot 2 \cdot a^2) - 8 + 12a - 6a^2 \)
\( = a^3 + (8 - a^3 - 12a + 6a^2) - 8 + 12a - 6a^2 \)
\( = a^3 - a^3 + 8 - 8 - 12a + 12a + 6a^2 - 6a^2 \)
\( = 0 \)
\( = \) R. H.S.
In simple words: This problem asks to prove an identity given a relationship between \( p \) and \( a \). The solution involves substituting \( p = 2-a \) into the left-hand side of the equation and simplifying the expression. Alternatively, recognizing \( a+p-2=0 \) allows applying the identity \( x^3+y^3+z^3=3xyz \) when \( x+y+z=0 \).

🎯 Exam Tip: For conditional proofs, direct substitution is a reliable method. Alternatively, if a sum of three terms is zero, remember the identity \( a^3+b^3+c^3=3abc \). This can often simplify the proof significantly.

Question 17. यदि x = 2y+6, तब सिद्ध कीजिए कि x3 - 8y3 - 36xy - 216 =0 .
Answer:
L.H.S. \( = x^3 - 8y^3 - 36xy - 216 \)
\( = (x)^3 - (2y)^3 - (6)^3 - 3(x)(2y)(6) \)
We are given \( x = 2y + 6 \)
\( \implies x - 2y = 6 \)
Consider 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 \( a^3+b^3+c^3 = 3abc \).
Here, let \( a = x, b = -2y, c = -6 \).
Then \( a+b+c = x - 2y - 6 \). Since \( x - 2y = 6 \), \( a+b+c = 6 - 6 = 0 \).
So, \( x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6) \)
\( x^3 - 8y^3 - 216 = 36xy \)
\( x^3 - 8y^3 - 36xy - 216 = 0 \)
\( = \) R.H.S.
In simple words: This proof leverages the algebraic identity \( a^3+b^3+c^3=3abc \) if \( a+b+c=0 \). By rearranging the given condition \( x = 2y+6 \) to \( x - 2y - 6 = 0 \), we identify \( a=x, b=-2y, c=-6 \) which satisfies the sum condition.

🎯 Exam Tip: Look for opportunities to apply the \( a^3+b^3+c^3=3abc \) identity. If you can manipulate the given condition to make \( a+b+c=0 \), the proof becomes much simpler. Be careful with signs when identifying \( a, b, c \).

Question 18. यदि x + y = -4, तब सिद्ध कीजिए कि x + y³ - 12xy + 64 = 0
Answer:
L.H.S. \( = x^3 + y^3 - 12xy + 64 \)
Given \( x + y = -4 \)
\( \implies x + y + 4 = 0 \)
Using the identity: if \( a+b+c=0 \), then \( a^3+b^3+c^3=3abc \).
Let \( a=x, b=y, c=4 \). Since \( a+b+c = x+y+4 = 0 \),
\( x^3 + y^3 + (4)^3 = 3(x)(y)(4) \)
\( x^3 + y^3 + 64 = 12xy \)
\( x^3 + y^3 - 12xy + 64 = 0 \)
\( = \) R.H.S.
In simple words: This proof uses the identity \( a^3+b^3+c^3=3abc \) when \( a+b+c=0 \). By transforming \( x+y=-4 \) into \( x+y+4=0 \), we can set \( a=x, b=y, c=4 \) and directly apply the identity to simplify the expression.

🎯 Exam Tip: When given a sum of two variables, check if a third constant can be introduced to make their sum zero. This allows for a direct application of the powerful \( a^3+b^3+c^3=3abc \) identity, simplifying cubic expressions efficiently.

Question 19. यदि a2 + b2 + c2 = 250 और ab+ bc + ca = 3, तब सिद्ध कीजिए कि a + b+ c = ±16
Answer:
सूत्र \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \).
\( (a + b + c)^2 = 250 + 2 \times 3 \)
\( (a + b + c)^2 = 250 + 6 \)
\( (a + b + c)^2 = 256 \)

\( (a + b + c) = \sqrt{256} = \pm 16 \)
In simple words: This problem uses the identity for the square of a trinomial: \( (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \). By substituting the given values for \( a^2+b^2+c^2 \) and \( ab+bc+ca \), we can find the value of \( (a+b+c)^2 \) and then take its square root.

🎯 Exam Tip: Memorize the expansion of \( (a+b+c)^2 \). This identity is frequently used to find the value of \( (a+b+c) \) when sums of squares and pairwise products are given. Remember to include both positive and negative roots when taking a square root.

Question 20. सिद्ध कीजिए कि
Answer:
माना \( a = 0.013 \) तथा \( b = 0.007 \)
\( = a + b = 0.013 + 0.007 = 0.020 \)
In simple words: This question implicitly asks for the sum of two given decimal numbers. By assigning variables to the numbers, the sum is directly calculated.

🎯 Exam Tip: Carefully read the context for questions that appear to have missing details. Sometimes, the "solution" itself implies the operation or value being sought, such as a simple addition here.

Question 21. यदि a + b ≠ 0 तब सिद्ध कीजिए कि समीकरण a(x - a) = 2ab - b(x - b) का हल x = a + b होता है |
Answer:
हलः \( \therefore a(x - a) = 2ab - b(x - b) \)

\( \implies ax - a^2 = 2ab - bx + b^2 \)

\( \implies ax + bx = a^2 + 2ab + b^2 \)

\( \implies x(a + b) = (a + b)^2 \)
दिया है कि \( a + b \neq 0 \), इसलिए दोनों पक्षों को \( (a+b) \) से विभाजित कर सकते हैं।

\( \implies x = \frac{(a+b)^2}{(a+b)} \)

\( \implies x = a + b \)
अतः \( x = a + b \) होता है।
In simple words: To prove the solution for \( x \), expand both sides of the given equation. Collect terms involving \( x \) on one side and constant terms on the other. Factor out \( x \) and then divide by \( (a+b) \) (since \( a+b \neq 0 \)) to isolate \( x \).

🎯 Exam Tip: When solving equations for a variable, expand expressions, group like terms, and apply algebraic rules to isolate the variable. Pay attention to conditions like \( a+b \neq 0 \) as they justify division by that term.

Question 22. सिद्ध कीजिए कि 2(a² + b²) = (a + b)2 तब a = b
Answer:
\( \therefore 2(a^2 + b^2) = (a + b)^2 \)

\( \implies 2a^2 + 2b^2 = a^2 + b^2 + 2ab \)

\( \implies 2a^2 - a^2 + 2b^2 - b^2 - 2ab = 0 \)

\( \implies a^2 + b^2 - 2ab = 0 \)

\( \implies (a - b)^2 = 0 \)

\( \implies a - b = 0 \)

\( \implies a = b \)
In simple words: This proof starts by expanding both sides of the given equation. Rearrange all terms to one side, which will form a perfect square trinomial \( (a-b)^2 \). Setting this to zero directly implies \( a=b \).

🎯 Exam Tip: For conditional proofs, expand and simplify the given equation. Recognizing perfect square trinomials like \( a^2 - 2ab + b^2 = (a-b)^2 \) is key to quickly reaching the desired conclusion.

Question 23. सिद्ध कीजिए कि 1 - x² + 2xy - y² = (1 + x - y)(1 - x + y)
Answer:
L.H.S. \( = 1 - x^2 + 2xy - y^2 \)
\( = 1 - (x^2 - 2xy + y^2) \)
\( = 1^2 - (x - y)^2 \)
\( = (1 - (x - y))(1 + (x - y)) \)
\( = (1 - x + y)(1 + x - y) \)
\( = \) R.H.S.
In simple words: This proof involves factoring by grouping. Rearrange the terms to group \( -x^2 + 2xy - y^2 \) as \( -(x^2 - 2xy + y^2) \), which is a perfect square \( -(x-y)^2 \). Then apply the difference of squares identity \( a^2 - b^2 = (a-b)(a+b) \).

🎯 Exam Tip: When factoring expressions with four terms, look for patterns that suggest difference of squares or perfect square trinomials. Grouping terms strategically can simplify the expression into a recognizable identity.

Question 24.
Answer:
\( x^2 + \frac{1}{x^2} = 14 \)

\( \implies \left(x + \frac{1}{x}\right)^2 - 2 = 14 \)

\( \implies \left(x + \frac{1}{x}\right)^2 = 14 + 2 \)

\( \implies \left(x + \frac{1}{x}\right)^2 = 16 \)

\( \implies x + \frac{1}{x} = 4 \)
घन करने पर,
\( \left(x + \frac{1}{x}\right)^3 = (4)^3 \)

\( \implies x^3 + \frac{1}{x^3} + 3 \cdot x \cdot \frac{1}{x} \left(x + \frac{1}{x}\right) = 64 \)

\( \implies x^3 + \frac{1}{x^3} + 3 (4) = 64 \)

\( \implies x^3 + \frac{1}{x^3} + 12 = 64 \)

\( \implies x^3 + \frac{1}{x^3} = 64 - 12 \)

\( \implies x^3 + \frac{1}{x^3} = 52 \)
In simple words: This problem involves finding the value of \( x^3 + \frac{1}{x^3} \) given \( x^2 + \frac{1}{x^2} = 14 \). First, find \( x + \frac{1}{x} \) using the identity \( \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \). Then, cube both sides of \( x + \frac{1}{x} \) and use the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \) to find the required expression.

🎯 Exam Tip: Remember the reciprocal identities: \( \left(x \pm \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} \pm 2 \) and \( \left(x \pm \frac{1}{x}\right)^3 = x^3 \pm \frac{1}{x^3} \pm 3\left(x \pm \frac{1}{x}\right) \). These are very useful for solving such problems efficiently.

Question 25.
Answer:
\( x + \frac{1}{x} = 3 \)
वर्ग करने पर,
\( \left(x + \frac{1}{x}\right)^2 = (3)^2 \)

\( \implies x^2 + \frac{1}{x^2} + 2 = 9 \)

\( \implies x^2 + \frac{1}{x^2} = 9 - 2 \)

\( \implies x^2 + \frac{1}{x^2} = 7 \)
पुनः वर्ग करने पर,
\( \left(x^2 + \frac{1}{x^2}\right)^2 = (7)^2 \)

\( \implies x^4 + \frac{1}{x^4} + 2 = 49 \)

\( \implies x^4 + \frac{1}{x^4} = 49 - 2 \)

\( \implies x^4 + \frac{1}{x^4} = 47 \)
In simple words: This problem demonstrates how to find higher powers of \( x + \frac{1}{x} \) type expressions. Given \( x + \frac{1}{x} = 3 \), square both sides to find \( x^2 + \frac{1}{x^2} \), then square again to find \( x^4 + \frac{1}{x^4} \).

🎯 Exam Tip: For expressions involving \( x + \frac{1}{x} \), repeatedly squaring the expression is an efficient way to find \( x^2 + \frac{1}{x^2} \), \( x^4 + \frac{1}{x^4} \), and so on. Remember the identity \( \left(A + \frac{1}{A}\right)^2 = A^2 + \frac{1}{A^2} + 2 \).

Question 26.
Answer:
\( x^2 + \frac{1}{x^2} = 34 \)

\( \implies \left(x + \frac{1}{x}\right)^2 - 2 = 34 \)

\( \implies \left(x + \frac{1}{x}\right)^2 = 36 \)

\( \implies x + \frac{1}{x} = 6 \)
घन करने पर,
\( \left(x + \frac{1}{x}\right)^3 = (6)^3 \)

\( \implies x^3 + \frac{1}{x^3} + 3 \cdot x \cdot \frac{1}{x} \left(x + \frac{1}{x}\right) = 216 \)

\( \implies x^3 + \frac{1}{x^3} + 3 (6) = 216 \)

\( \implies x^3 + \frac{1}{x^3} + 18 = 216 \)

\( \implies x^3 + \frac{1}{x^3} = 216 - 18 \)

\( \implies x^3 + \frac{1}{x^3} = 198 \)
In simple words: Given \( x^2 + \frac{1}{x^2} = 34 \), this problem asks to find \( x^3 + \frac{1}{x^3} \). First, calculate \( x + \frac{1}{x} \) using the identity for \( (A+B)^2 \). Then, cube both sides of the result and apply the identity for \( (A+B)^3 \) to find the required cubic expression.

🎯 Exam Tip: This type of problem often appears. Always break it down: first find \( x \pm \frac{1}{x} \) from the squared term, and then use that to find the cubic term. Be careful with signs if the expression involves subtraction, i.e., \( x - \frac{1}{x} \).