Question

What is the complete factorisation of 32x^8 - 2y^8

Answer 1

Let us denote the polynomial by $T$.

Then $T=32x^8-2y^8$.

$\implies \dfrac{T}{2} = 16x^8 - y^8$

$\implies \dfrac{T}{2} = (4x^4)^2 - (y^4)^2$

$\implies \dfrac{T}{2} = ( 4x^4 + y^4 ) ( 4x^4 - y^4 )$

$\implies \dfrac{T}{2} = ( 4x^4 + y^4 ) ( 2x^2 + y^2 ) ( 2x^2 - y^2 )$

$\therefore T = 2 ( 4x^4 + y^4 ) ( 2x^2 + y^2 ) ( 2x^2 - y^2 )$