Question:
Factorise
$x^{6}-7 x^{3}-8$
Solution:
Let $x^{3}=y$
So, the equation becomes
$y^{2}-7 y-8=y^{2}-8 y+y-8$
$=y(y-8)+(y-8)$
$=(y-8)(y+1)$
$=\left(x^{3}-8\right)\left(x^{3}+1\right)$
$=(x-2)\left(x^{2}+4+2 x\right)(x+1)\left(x^{2}+1-x\right)$
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