Factorize :
$x^{6} - 7 x^{3} - 8$

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Solution

given, $x^{6} - 7 x^{3} - 8$

$= y^{2} - 7 y - 8$ , where $y = x^{3}$

$= y^{2} - 8 y + y - 8$

$= (y^{2} - 8 y) + (y - 8)$

$= y (y - 8) + 1 (y - 8)$

$= (y - 8) (y + 1)$

$= (x^{3} - 8) (x^{3} + 1)$

$= (x^{3} - 2^{3}) (x^{3} + 1^{3})$

$= (x - 2) (x^{2} + 2 x + 4) (x + 1) (x^{2} - x + 1)$

$= (x - 2) (x + 1) (x^{2} + 2 x + 4) (x^{2} - x + 1)$

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