The correct option is A (2a/3 b/2)[4a^2/9+b^2/4+ab/3] 8a^3/27 b^3/8 =(2a/3)^3 (b/2)^3 We know that, x^3 y^3 = (x y)(x^2+xy+y^2) = (2a/3 b/2)[4a^2/9+b^2/4+ab ...

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Factorise 83/...

Question

Factorise $\frac{8 a^{3}}{27} - \frac{b^{3}}{8}$

(\frac{2 a}{3} - \frac{b}{2}) [\frac{4 a^{2}}{9} + \frac{b^{2}}{4} + \frac{a b}{3}]

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(\frac{2 a}{3} - \frac{b^{3}}{2}) [\frac{4 a}{9} + \frac{b^{2}}{4} + \frac{a b}{3}]

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(\frac{2 a}{3} - \frac{b^{2}}{2}) [\frac{4 a^{2}}{9} + \frac{b^{3}}{4} + \frac{a b}{3}]

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(\frac{2 a}{3} - \frac{b}{2}) [\frac{4 a^{2}}{9} + \frac{b^{2}}{4} + \frac{a b^{2}}{3}]

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Solution

The correct option is A $(\frac{2 a}{3} - \frac{b}{2}) [\frac{4 a^{2}}{9} + \frac{b^{2}}{4} + \frac{a b}{3}]$

$\frac{8 a^{3}}{27} - \frac{b^{3}}{8}$
$= (\frac{2 a}{3})^{3} - (\frac{b}{2})^{3}$
We know that,
{ $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ }
$= (\frac{2 a}{3} - \frac{b}{2}) [\frac{4 a^{2}}{9} + \frac{b^{2}}{4} + \frac{a b}{3}]$

Suggest Corrections

Factorise 8a327−b38

The correct option is A (2a3−b2)[4a29+b24+ab3] 8a327−b38 =(2a3)3−(b2)3 We know that, { x3−y3=(x−y)(x2+xy+y2) } =(2a3−b2)[4a29+b24+ab3]

Factorise $\frac{8 a^{3}}{27} - \frac{b^{3}}{8}$