Factorise the following- 8y3 - 125-y3

The correct option is B (2y 5/y) ( 4y^2+10+25/y^2) 8y^3 125y^3 =(2y)^3 (5y)^3 Using, a^3 b^3=(a b)(a^2+ab+b^2) =(2y 5y)[(2y)^2+(2y ...

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Standard IX

Mathematics

Algebraic Identities

Factorise the...

Question

Factorise the following: $8 y^{3} - \frac{125}{y^{3}}$

(2 y + \frac{5}{y}) (y^{2} + 10 + \frac{25}{y^{2}})

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(2 y - \frac{5}{y}) (4 y^{2} + 10 + \frac{25}{y^{2}})

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(2 y - \frac{5}{y}) (4 y^{2} + \frac{25}{y^{2}})

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(y - \frac{5}{y}) (4 y^{2} + 10 + \frac{25}{y})

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Solution

The correct option is B $(2 y - \frac{5}{y}) (4 y^{2} + 10 + \frac{25}{y^{2}})$
$8 y^{3} - \frac{125}{y^{3}}$
$= (2 y)^{3} - (\frac{5}{y})^{3}$
Using, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$
$= (2 y - \frac{5}{y}) [(2 y)^{2} + (2 y) (\frac{5}{y}) + (\frac{5}{y})^{2}]$
$= (2 y - \frac{5}{y}) (4 y^{2} + 10 + \frac{25}{y^{2}})$

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Factorise the following: 8y3−125y3

The correct option is B (2y−5y)(4y2+10+25y2)8y3−125y3=(2y)3−(5y)3Using, a3−b3=(a−b)(a2+ab+b2)=(2y−5y)[(2y)2+(2y)(5y)+(5y)2]=(2y−5y)(4y2+10+25y2)

Factorise the following: $8 y^{3} - \frac{125}{y^{3}}$