Maharashtra Board Class 8 Maths Chapter 6 Factorisation of Algebraic Expressions Set 6.4 Solutions

Question 1. Simplify:
(i) \( \frac{m^2-n^2}{(m+n)^2} \times \frac{m^2+mn+n^2}{m^3-n^3} \)
(ii) \( \frac{a^2+10a+21}{a^2+6a-7} \times \frac{a^2-1}{a+3} \)
(iii) \( \frac{8x^3-27y^3}{4x^2-9y^2} \)
(iv) \( \frac{x^2-5x-24}{(x+3)(x+8)} \times \frac{x^2-64}{(x-8)^2} \)
(v) \( \frac{3x^2-x-2}{x^2-7x+12} \div \frac{3x^2-7x-6}{x^2-4} \)
(vi) \( \frac{4x^2-11x+6}{16x^2-9} \)
(vii) \( \frac{a^3-27}{5a^2-16a+3} \div \frac{a^2+3a+9}{25a^2-1} \)
(viii) \( \frac{1-2x+x^2}{1-x^3} \times \frac{1+x+x^2}{1+x} \)
Answer:
(i)
\( = \frac{m^2-n^2}{(m+n)^2} \times \frac{m^2+mn+n^2}{m^3-n^3} \)
\( = \frac{(m+n)(m-n)}{(m+n)(m+n)} \times \frac{m^2+mn+n^2}{(m-n)(m^2+mn+n^2)} \)
\( = \frac{1}{m+n} \)

(ii)
\( = \frac{a^2+10a+21}{a^2+6a-7} \times \frac{a^2-1}{a+3} \)
\( = \frac{a^2+7a+3a+21}{a^2+7a-a-7} \times \frac{a^2-1^2}{a+3} \)
\( = \frac{a(a+7)+3(a+7)}{a(a+7)-1(a+7)} \times \frac{(a+1)(a-1)}{a+3} \)
\( = \frac{(a+7)(a+3)}{(a+7)(a-1)} \times \frac{(a+1)(a-1)}{a+3} \)
\( = a+1 \)

(iii)
\( = \frac{8x^3-27y^3}{4x^2-9y^2} \)
\( = \frac{(2x)^3-(3y)^3}{(2x)^2-(3y)^2} \)
\( = \frac{(2x-3y)((2x)^2+(2x)(3y)+(3y)^2)}{(2x+3y)(2x-3y)} \)
\( = \frac{(2x)^2+(2x)(3y)+(3y)^2}{2x+3y} \)
\( = \frac{4x^2+6xy+9y^2}{2x+3y} \)

(iv)
\( = \frac{x^2-5x-24}{(x+3)(x+8)} \times \frac{x^2-64}{(x-8)^2} \)
\( = \frac{x^2-8x+3x-24}{(x+3)(x+8)} \times \frac{x^2-8^2}{(x-8)^2} \)
\( = \frac{x(x-8)+3(x-8)}{(x+3)(x+8)} \times \frac{(x+8)(x-8)}{(x-8)(x-8)} \)
\( = \frac{(x-8)(x+3)}{(x+3)(x+8)} \times \frac{(x+8)(x-8)}{(x-8)(x-8)} \)
\( = 1 \)

(v)
\( = \frac{3x^2-x-2}{x^2-7x+12} \div \frac{3x^2-7x-6}{x^2-4} \)
\( = \frac{3x^2-x-2}{x^2-7x+12} \times \frac{x^2-4}{3x^2-7x-6} \)
\( = \frac{3x^2-3x+2x-2}{x^2-4x-3x+12} \times \frac{x^2-2^2}{3x^2-9x+2x-6} \)
\( = \frac{3x(x-1)+2(x-1)}{x(x-4)-3(x-4)} \times \frac{(x+2)(x-2)}{3x(x-3)+2(x-3)} \)
\( = \frac{(x-1)(3x+2)}{(x-4)(x-3)} \times \frac{(x+2)(x-2)}{(x-3)(3x+2)} \)
\( = \frac{(x-1)(x-2)(x+2)}{(x-3)(x-4)} \)

(vi)
\( = \frac{4x^2-11x+6}{16x^2-9} \)
\( = \frac{4x^2-8x-3x+6}{(4x)^2-3^2} \)
\( = \frac{4x(x-2)-3(x-2)}{(4x+3)(4x-3)} \)
\( = \frac{(x-2)(4x-3)}{(4x+3)(4x-3)} \)
\( = \frac{x-2}{4x+3} \)

(vii)
\( = \frac{a^3-27}{5a^2-16a+3} \div \frac{a^2+3a+9}{25a^2-1} \)
\( = \frac{a^3-27}{5a^2-16a+3} \times \frac{25a^2-1}{a^2+3a+9} \)
\( = \frac{a^3-3^3}{5a^2-15a-a+3} \times \frac{(5a)^2-1^2}{a^2+3a+9} \)
\( = \frac{(a-3)(a^2+3a+9)}{5a(a-3)-1(a-3)} \times \frac{(5a+1)(5a-1)}{a^2+3a+9} \)
\( = \frac{(a-3)(a^2+3a+9)}{(a-3)(5a-1)} \times \frac{(5a+1)(5a-1)}{a^2+3a+9} \)
\( = 5a+1 \)

(viii)
\( = \frac{1-2x+x^2}{1-x^3} \times \frac{1+x+x^2}{1+x} \)
\( = \frac{(1-x)^2}{(1-x)(1+x+x^2)} \times \frac{1+x+x^2}{1+x} \)
\( = \frac{(1-x)(1-x)}{(1-x)(1+x+x^2)} \times \frac{1+x+x^2}{1+x} \)
\( = \frac{1-x}{1+x} \)
In simple words: This question requires simplifying various algebraic expressions by applying factorization techniques for quadratic, cubic, and difference of squares/cubes polynomials, followed by canceling common factors. Each sub-part involves a different combination of these techniques.

🎯 Exam Tip: Always look for common factors and apply appropriate factorization formulas (like \(a^2-b^2\), \(a^3-b^3\), and quadratic factorization) to simplify expressions. Pay close attention to signs and exponents during each step to avoid errors.