Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Set 5.3 Solutions

Question 1. Expand:
(i) \((2m – 5)^3\)
(ii) \((4 – p)^3\)
(iii) \((7x – 9y)^3\)
(iv) \((58)^3\)
(v) \((198)^3\)
(vi) \((2p - \frac{1}{2p})^3\)
(vii) \((1 - \frac{1}{a})^3\)
(viii) \((\frac{x}{3} - \frac{3}{x})^3\)
Answer:
(i) Here, \(a = 2m\) and \(b = 5\) \[(2m – 5)^3\] \[= (2m)^3 – 3(2m)^2 (5) + 3(2m) (5)^2 – (5)^3\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3]\] \[= 8m^3 – 3(4m^2)(5) + 3(2m)(25) – 125\] \[= 8m^3 – 60m^2 + 150m – 125\]
(ii) Here, \(a = 4\) and \(b = p\) \[(4 – p)^3 = (4)^3 – 3(4)^2(p) + 3(4)(p)^2 – (p)^3\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3]\] \[= 64 – 3(16)(p) + 3(4)(p^2) – p^3\] \[= 64 - 48p + 12p^2 – p^3\]
(iii) Here, \(a = 7x\) and \(b = 9y\) \[(7x – 9y)^3\] \[= (7x)^3 – 3(7x)^2 (9y) + 3 (7x)(9y)^2 – (9y)^3\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3]\] \[= 343x^3 – 3(49x^2)(9y) + 3(7x)(81y^2) – 729y^3\] \[= 343x^3 – 1323x^2y + 1701xy^2 – 729y^3\]
(iv) \((58)^3 = (60 – 2)^3\) Here, \(a = 60\) and \(b = 2\) \[(58)^3 = (60)^3 – 3(60)^2(2) + 3(60)(2)^2 – (2)^3\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3]\] \[= 216000 – 3(3600)(2) + 3(60)(4) – 8\] \[= 216000 – 21600 + 720 – 8\] \[=195112\]
(v) \((198)^3 = (200 – 2)^3\) Here, \(a = 200\) and \(b = 2\) \[(198)^3 = (200)^3 – 3(200)^2(2) + 3(200)(2)^2 – (2)^3\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3]\] \[= 8000000 – 3(40000)(2) + 3(200)(4) – 8\] \[= 8000000 – 240000 + 2400 – 8\] \[= 7762392\]
(vi) Here, \(a = 2p\) and \(b = \frac{1}{2p}\) \[(2p - \frac{1}{2p})^3\] \[= (2p)^3 - 3(2p)^2 (\frac{1}{2p}) + 3(2p) (\frac{1}{2p})^2 - (\frac{1}{2p})^3\] \[\dots[(a - b)^3 = a^3-3a^2b + 3ab^2 - b^3]\] \[= 8p^3 - 3(2p) (\frac{1}{2p}) + 3(2p) (\frac{1}{2p})^2 - \frac{1}{8p^3}\] \[= 8p^3 - 3(2p) + 3(\frac{1}{2p}) - \frac{1}{8p^3}\] \[= 8p^3 – 6p + \frac{3}{2p} - \frac{1}{8p^3}\]
(vii) Here, \(A = 1\) and \(B = \frac{1}{a}\) \[(1 - \frac{1}{a})^3 = (1)^3 - 3(1)^2 (\frac{1}{a}) + 3(1)(\frac{1}{a})^2 - (\frac{1}{a})^3\] \[\dots[(a - b)^3 = a^3-3a^2b + 3ab^2 - b^3]\] \[= 1 - \frac{3}{a} + \frac{3}{a^2} - \frac{1}{a^3}\]
(viii) Here, \(a = \frac{x}{3}\) and \(b = \frac{3}{x}\) \[(\frac{x}{3} - \frac{3}{x})^3\] \[= (\frac{x}{3})^3 - 3(\frac{x}{3})^2 (\frac{3}{x}) + 3(\frac{x}{3}) (\frac{3}{x})^2 - (\frac{3}{x})^3\] \[\dots[(a - b)^3 = a^3-3a^2b + 3ab^2 - b^3]\] \[= \frac{x^3}{27} - 3(\frac{x^2}{9}) (\frac{3}{x}) + 3(\frac{x}{3}) (\frac{9}{x^2}) - \frac{27}{x^3}\] \[= \frac{x^3}{27} - x + \frac{9}{x} - \frac{27}{x^3}\] In simple words: This question asks us to expand various algebraic expressions using the formula for the cube of a binomial, \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\). For numerical problems like \((58)^3\), we rewrite them as a binomial, e.g., \((60-2)^3\), and then apply the same expansion formula.

🎯 Exam Tip: Remember to correctly identify 'a' and 'b' in each expression and pay close attention to signs, especially when squaring or cubing negative terms, to avoid calculation errors.

Question 2. Simplify:
(i) \((2a + b)^3 – (2a – b)^3\)
(ii) \((3r – 2k)^3 + (3r + 2k)^3\)
(iii) \((4a – 3)^3 – (4a + 3)^3\)
(iv) \((5x – 7y)^3 + (5x + 7y)^3\)
Answer:
(i) \((2a + b)^3 – (2a - b)^3\) \[= [(2a)^3 + 3(2a)^2(b) + 3 (2a)(b)^2 + (b)^3] – [(2a)^3 – 3(2a)^2(b) + 3 (2a)(b)^2 – (b)^3]\] \[\dots[(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, (a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3]\] \[= (8a^3 + 12a^2b + 6ab^2 + b^3) – (8a^3 – 12a^2b + 6ab^2 – b^3)\] \[= 8a^3 + 12a^2b + 6ab^2 + b^3 – 8a^3 + 12a^2b – 6ab^2 + b^3\] \[= 8a^3 – 8a^3 + 12a^2b + 12a^2b + 6ab^2 – 6ab^2 + b^3 + b^3\] \[= 24a^2b + 2b^3\]
(ii) \((3r – 2k)^3 + (3r + 2k)^3\) \[= [(3r)^3 – 3(3r)^2(2k) + 3(3r)(2k)^2 – (2k)^3] + [(3r)^3 + 3(3r)^2(2k) + 3(3r)(2k)^2 + (2k)^3]\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3, (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3]\] \[= (27r^3 – 54r^2k + 36rk^2 – 8k^3) + (27r^3 + 54r^2k + 36rk^2 + 8k^3)\] \[= 27r^3 – 54r^2k + 36rk^2 – 8k^3 + 27r^3 + 54r^2k + 36rk^2 + 8k^3\] \[= 27r^3 + 27r^3 – 54r^2k + 54r^2k + 36rk^2 + 36rk^2 – 8k^3 + 8k^3\] \[= 54r^3 + 72rk^2\]
(iii) \((4a – 3)^3 – (4a + 3)^3\) \[= [(4a)^3 – 3(4a)^2 (3) + 3(4a)(3)^2 – (3)^3] – [(4a)^3 + 3(4a)^2(3) + 3(4a)(3)^2 + (3)^3]\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3, (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3]\] \[= (64a^3 – 144a^2 + 108a – 27) – (64a^3 + 144a^2 + 108a + 27)\] \[= 64a^3 – 144a^2 + 108a – 27 – 64a^3 -144a^2 – 108a – 27\] \[= 64a^3 – 64a^3 – 144a^2 – 144a^2 + 108a – 108a – 27 – 27\] \[= -288a^2 - 54\]
(iv) \((5x – 7y)^3 + (5x + 7y)^3\) \[= [(5x)^3 – 3(5x)^2(7y) + 3(5x)(7y)^2 – (7y)^3] + [(5x)^3 + 3(5x)^2 (7y) + 3(5x) (7y)^2 +(7y)^3]\] \[\dots[(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3, (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3]\] \[= (125x^3 – 525x^2y + 735xy^2 – 343y^3) + (125x^3 + 525x^2y + 735xy^2 + 343y^3)\] \[= 125x^3 + 125x^3 – 525x^2y + 525x^2y + 735xy^2 + 735xy^2 – 343y^3 + 343y^3\] \[= 250x^3 + 1470xy^2\] In simple words: This question involves simplifying expressions that are sums or differences of two binomial cubes. We use the expansion formulas for \((a+b)^3\) and \((a-b)^3\), then combine like terms and cancel out terms with opposite signs to reach the simplified form.

🎯 Exam Tip: When simplifying sums or differences of cubic expansions, pay close attention to distributing the negative sign correctly if it's a subtraction. This is a common source of error.

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.3 Intext Questions And Activities

Question 1. Make two cubes of side a and of side b each. Make six parallelopipeds; three of them measuring a × a × b and the remaining three measuring b × b × a. Arrange all these solid figures properly and make a cube of side (a + b). (Textbook pg. no. 25)
Answer: Solution: \[(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\] \[=a \times a \times a + 3 \times a \times a \times b + 3 \times a \times b \times b + b \times b \times b\]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र \((a+b)^3\) के विस्तार को एक बड़े घन के निर्माण के माध्यम से दर्शाता है। इसमें एक बड़ा घन (भुजा 'a' का), तीन आयताकार प्रिज्म (आकार \(a \times a \times b\)), तीन अन्य आयताकार प्रिज्म (आकार \(a \times b \times b\)), और एक छोटा घन (भुजा 'b' का) शामिल हैं। इन सभी टुकड़ों को व्यवस्थित करके एक बड़ा घन बनाया जाता है जिसकी प्रत्येक भुजा की लंबाई \((a+b)\) होती है।

🎯 Exam Tip: Visualizing algebraic identities using geometric models, like the cube expansion shown here, can help deepen understanding and memory, especially for complex formulas.