Factorise:
$x^{6} - y^{6}$

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Solution

We know the identity $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$

Using the above identity, the equation $x^{6} - y^{6}$ can be factorised as follows:

$x^{6} - y^{6} = (x^{2})^{3} - {(y^{2})}^{3} = (x^{2} - y^{2}) [(x^{2})^{2} + {(y^{2})}^{2} + (x^{2} \times y^{2})] = (x - y) (x + y) (x^{2} + y^{2} + x y) (x^{2} + y^{2} - x y)$
(using identities $a^{2} - b^{2} = (a + b) (a - b)$ and $a^{4} + b^{4} + a^{2} b^{2} = (a^{2} + b^{2} + a b) (a^{2} + b^{2} - a b)$ )

Hence, $x^{6} - y^{6} = (x - y) (x + y) (x^{2} + y^{2} + x y) (x^{2} + y^{2} - x y)$

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Similar questions

x^{6} - y^{6}

Factorize $x^{6} + y^{6}$

\sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a (x^{3} - y^{3})

\frac{d y}{d x} = f (x, y) \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}

\sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a^{3} (x^{3} - y^{3})

\frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}

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