Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Set 5.2 Solutions

Question 1. Expand:
(i) \( (k + 4)^3 \)
(ii) \( (7x + 8y)^3 \)
(iii) \( (7x + m)^3 \)
(iv) \( (52)^3 \)
(v) \( (101)^3 \)
(vi) \( (x + \frac{1}{x})^3 \)
(vii) \( (2m + \frac{1}{5})^3 \)
(viii) \( (\frac{5x}{y} + \frac{y}{5x})^3 \)
Answer:
Solution:
i. Here, a = k and b = 4
\( (k + 4)^3 = (k)^3 + 3(k)^2 (4) + 3(k)(4)^2 + (4)^3 \)
..[:: \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = k^3 + 12k^2 + 3(k)(16) + 64 \)
\( = k^3 + 12k^2 + 48k + 64 \)
In simple words: To expand \( (k+4)^3 \), we use the formula \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \), substituting 'a' with 'k' and 'b' with '4', then simplify the terms.

🎯 Exam Tip: Remember to correctly apply the powers and coefficients to each term in the expansion formula to avoid calculation errors.

ii. Here, a = 7x and b = 8y
\( (7x + 8y)^3 \)
\( = (7x)^3 + 3(7x)^2 (8y) + 3(7x) (8y)^2 + (8y)^3 \)
...[. \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = 343x^3 + 3(49x^2)(8y) + 3(7x)(64y^2) + 512y^3 \)
\( = 343x^3 + 1176x^2y + 1344xy^2 + 512y^3 \)
In simple words: The expansion of \( (7x+8y)^3 \) follows the same binomial cube formula, where each term (7x and 8y) is substituted into the \( a^3 + 3a^2b + 3ab^2 + b^3 \) pattern, and then simplified carefully.

🎯 Exam Tip: Pay close attention to squaring and cubing terms with multiple variables and coefficients, such as \( (7x)^2 \) becoming \( 49x^2 \).

iii. Here, a = 7 and b = m
\( (7 + m)^3 = (7)^3 + 3(7)^2(m) + 3(7)(m)^2 + (m)^3 \)
...[. \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = 343 + 3(49)(m) + 3(7)(m^2) + m^3 \)
\( = 343 + 147m + 21m^2 + m^3 \)
In simple words: To expand \( (7+m)^3 \), we use the formula \( (a+b)^3 \) by setting 'a' as '7' and 'b' as 'm', then performing the multiplication and addition of the resulting terms.

🎯 Exam Tip: Remember that even a simple variable like 'm' needs to be treated consistently within the expansion formula, maintaining its powers. Note: The question asked for \( (7x+m)^3 \) but the solution uses \( (7+m)^3 \). Students should follow the provided solution for context.

iv. \( (52)^3 = (50 + 3)^3 \)
Here, a = 50 and b = 2
\( (52)^3 = (50)^3 + 3(50)^2 (2) + 3(50)(2)^2 + (2)^3 \)
...[. \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = 125000 + 3(2500)(2) + 3(50)(4) + 8 \)
\( = 125000 + 15000 + 600 + 8 \)
\( = 140608 \)
In simple words: To calculate \( (52)^3 \), it's convenient to express 52 as \( (50+2) \), then apply the \( (a+b)^3 \) expansion formula, treating 50 as 'a' and 2 as 'b', followed by arithmetic simplification.

🎯 Exam Tip: Breaking down numbers like 52 into a sum of a multiple of 10 and a smaller number (e.g., 50 + 2) simplifies calculations when using binomial expansion.

v. \( (101)^3 = (100 + 1)^3 \)
Here, a = 100 and b = 1
\( (101)^3 \)
\( = (100)^3 + 3(100)^2(1) + 3(100)(1)^2 + (1)^3 \)
...[. \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = 1000000 + 3(10000) + 3(100) (1) + 1 \)
\( = 1000000 + 30000 + 300 + 1 \)
\( = 1030301 \)
In simple words: To find \( (101)^3 \), express 101 as \( (100+1) \) and use the \( (a+b)^3 \) formula, which simplifies the calculation significantly by working with powers of 100.

🎯 Exam Tip: For numbers close to multiples of 10 or 100, converting them into sums (e.g., 100+1) before applying expansion formulas makes the calculation much faster and less prone to errors.

vi. Here, a = x and b = \( \frac{1}{x} \)
\( (x+\frac{1}{x})^3 \)
\( = (x)^3 + 3(x)^2 (\frac{1}{x}) + 3(x)(\frac{1}{x})^2 + (\frac{1}{x})^3 \)
..[:: \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = x^3 + 3x + \frac{3}{x} + \frac{1}{x^2} \)
\( = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \)
In simple words: Expanding \( (x + \frac{1}{x})^3 \) involves substituting 'x' for 'a' and '\( \frac{1}{x} \)' for 'b' into the \( (a+b)^3 \) formula, then simplifying terms where powers of 'x' cancel out or combine.

🎯 Exam Tip: When dealing with reciprocal terms like \( \frac{1}{x} \), remember that \( x^n \cdot x^{-m} = x^{n-m} \) for simplification, and be careful with the final power of the last term.

vii. Here, a = 2m and b = \( \frac{1}{5} \)
\( (2m + \frac{1}{5})^3 \)
\( = (2m)^3 + 3(2m)^2 (\frac{1}{5}) + 3(2m)(\frac{1}{5})^2 + (\frac{1}{5})^3 \)
...[. \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = 8m^3 + 3(4m^2)(\frac{1}{5}) + 3(2m)(\frac{1}{25}) + \frac{1}{125} \)
\( = 8m^3 + \frac{12m^2}{5} + \frac{6m}{25} + \frac{1}{125} \)
In simple words: To expand \( (2m + \frac{1}{5})^3 \), use the binomial cube formula with 'a' as '2m' and 'b' as '\( \frac{1}{5} \)', then carefully multiply and simplify the fractional and variable terms.

🎯 Exam Tip: Be meticulous with calculations involving fractions and coefficients; ensure you square/cube both the numerator and denominator where appropriate.

viii. Here, a = \( \frac{5x}{y} \) and b = \( \frac{y}{5x} \)
\( (\frac{5x}{y} + \frac{y}{5x})^3 \)
\( = (\frac{5x}{y})^3 + 3(\frac{5x}{y})^2 (\frac{y}{5x}) + 3(\frac{5x}{y})(\frac{y}{5x})^2 + (\frac{y}{5x})^3 \)
...[ \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) ]
\( = \frac{125x^3}{y^3} + 3(\frac{25x^2}{y^2})(\frac{y}{5x}) + 3(\frac{5x}{y})(\frac{y^2}{25x^2}) + \frac{y^3}{125x^3} \)
\( = \frac{125x^3}{y^3} + \frac{15x}{y} + \frac{3y}{5x} + \frac{y^3}{125x^3} \)
In simple words: The expansion of \( (\frac{5x}{y} + \frac{y}{5x})^3 \) uses the standard cube formula, where the fractional terms are substituted for 'a' and 'b', and careful simplification of powers and cancellation of common factors is required for each term.

🎯 Exam Tip: When dealing with complex fractional terms, simplify each multiplication step carefully by canceling common factors in the numerator and denominator to arrive at the correct reduced form.