SSC (English Medium) Class 8Maharashtra State Board
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# Simplify : 8 X 3 − 27 Y 3 4 X 2 − 9 Y 2 - SSC (English Medium) Class 8 - Mathematics

#### Question

Simplify :

$\frac{8 x^3 - 27 y^3}{4 x^2 - 9 y^2}$

#### Solution

It is known that,

$a^2 - b^2 = \left( a + b \right)\left( a - b \right) ; a^3 - b^3 = \left( a - b \right)\left( a^2 + ab + b^2 \right)$

$\ \frac{8 x^3 - 27 y^3}{4 x^2 - 9 y^2}$

$= \frac{\left( 2x \right)^3 - \left( 3y \right)^3}{\left( 2x \right)^2 - \left( 3y \right)^2}$

$= \frac{\left( 2x - 3y \right)\left[ \left( 2x \right)^2 + \left( 2x \right) \times \left( 3y \right) + \left( 3y \right)^2 \right]}{\left( 2x + 3y \right)\left( 2x - 3y \right)}$

$= \frac{\left( 2x - 3y \right)\left( 4 x^2 + 6xy + 9 y^2 \right)}{\left( 2x + 3y \right)\left( 2x - 3y \right)}$

$= \frac{4 x^2 + 6xy + 9 y^2}{\left( 2x + 3y \right)}$

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 6: Factorisation of Algebraic expressions
Practice Set 6.4 | Q: 3 | Page no. 33
Solution Simplify : 8 X 3 − 27 Y 3 4 X 2 − 9 Y 2 Concept: Factors of A3 - B3.
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