Factorise

Let $x^{3}=y$

So, the equation becomes 

$y^{2}-7 y-8=y^{2}-8 y+y-8$

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

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

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

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