The correct option is D (x y)(x+y)(x^2+y^2)(x^2 √(2)xy+y^2)(x^2+√(2)xy+y^2) x^8 y^8 =(x^4 y^4)(x^4+y^4) =(x^2 y^2)(x^2+y^2)(x^4+y^4) =( ...

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Visualisation of Factorisation

Factorise: ...

Question

Factorise: $x^{8} - y^{8}$

(x - y) (x + y) (x^{2} - y^{2}) (x^{2} + \sqrt{2} x y - y^{2}) (x^{2} + \sqrt{2} x y + y^{2})

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(x - y) (x + y) (x^{2} + y^{2}) (x^{2} - \sqrt{2} x y + y^{2}) (x^{2} - \sqrt{2} x y + y^{2})

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(x - y) (x + y) (x^{2} - y^{2}) (x^{2} + \sqrt{2} x y + y^{2}) (x^{2} + \sqrt{2} x y + y^{2})

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(x - y) (x + y) (x^{2} + y^{2}) (x^{2} - \sqrt{2} x y + y^{2}) (x^{2} + \sqrt{2} x y + y^{2})

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Solution

The correct option is D $(x - y) (x + y) (x^{2} + y^{2}) (x^{2} - \sqrt{2} x y + y^{2}) (x^{2} + \sqrt{2} x y + y^{2})$
$x^{8} - y^{8}$

$= (x^{4} - y^{4}) (x^{4} + y^{4})$

$= (x^{2} - y^{2}) (x^{2} + y^{2}) (x^{4} + y^{4})$

$= (x - y) (x + y) (x^{2} + y^{2}) [(x^{2})^{2} + (y^{2})^{2} + 2 x^{2} y^{2} - 2 x^{2} y^{2}]$

$= (x - y) (x + y) (x^{2} + y^{2}) [(x^{2} + y^{2})^{2} - (\sqrt{2} x y)^{2}]$

$= (x - y) (x + y) (x^{2} + y^{2}) [(x^{2} + y^{2}) - (\sqrt{2} x y)] [(x^{2} + y^{2}) + (\sqrt{2} x y)]$

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