# Factorise. Y 3 + 1 8 Y 3 - Mathematics

Sum

Factorise.

y^3 + 1/(8y^3)

#### Solution

It is known that,

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

$\ y^3 + \frac{1}{8 y^3}$

$= \left( y \right)^3 + \left( \frac{1}{2y} \right)^3$

$= \left( y + \frac{1}{2y} \right)\left\{ \left( y \right)^2 + \left( \frac{1}{2y} \right)^2 - \left( y \right) \times \left( \frac{1}{2y} \right) \right\}$

$= \left( y + \frac{1}{2y} \right)\left( y^2 + \frac{1}{4 y^2} - \frac{1}{2} \right)$

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#### APPEARS IN

Balbharati Mathematics 8th Standard Maharashtra State Board
Chapter 6 Factorisation of Algebraic expressions
Practice Set 6.2 | Q 6 | Page 31
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