Maharashtra State BoardSSC (English Medium) 8th Standard
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Simplify : 8 X 3 − 27 Y 3 4 X 2 − 9 Y 2 - Mathematics

Sum

Simplify :

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

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Solution

It is known that,

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

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

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

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

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

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

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APPEARS IN

Balbharati Mathematics 8th Standard Maharashtra State Board
Chapter 6 Factorisation of Algebraic expressions
Practice Set 6.4 | Q 3 | Page 33
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