Question
Factorise:
$x^6-7 x^3-8$
$x^6-7 x^3-8$
✓
Answer
Given equation is $x^6-7 x^3-8$.
Putting $x^3=y$, we get
$y^2-7 y-8$
$=y^2-8 y+y-8$
$=y(y-8)+1(y-8)$
$=(y-8)(y+1)$
$=\left(x^3-8\right)\left(x^3+1\right)$
$=\left(x^3-2^3\right)\left(x^3+1^3\right)$
$=(x-2)\left(x^2+2 x+4\right)(x+1)\left(x^2-x+1\right)$
$=(x-2)(x+1)\left(x^2+2 x+4\right)\left(x^2-x+1\right)$
Putting $x^3=y$, we get
$y^2-7 y-8$
$=y^2-8 y+y-8$
$=y(y-8)+1(y-8)$
$=(y-8)(y+1)$
$=\left(x^3-8\right)\left(x^3+1\right)$
$=\left(x^3-2^3\right)\left(x^3+1^3\right)$
$=(x-2)\left(x^2+2 x+4\right)(x+1)\left(x^2-x+1\right)$
$=(x-2)(x+1)\left(x^2+2 x+4\right)\left(x^2-x+1\right)$
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