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

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Solution

Consider, $x^{6} - 7 x^{3} - 8$
$= x^{6} - 8 x^{3} + x^{3} - 8$
$= x^{3} (x^{3} - 8) + x^{3} - 8$
$= (x^{3} + 1) (x^{3} - 8)$
$= (x^{3} + 1^{3}) (x^{3} - 2^{3})$
$= (x + 1) (x^{2} - x + 1) (x - 2) (x^{2} + 2 x + 4)$
Hence, $x^{6} - 7 x^{3} - 8 = (x + 1) (x^{2} - x + 1) (x - 2) (x^{2} + 2 x + 4)$

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