Question 622008
{{{x^8-1=(x^4)^2-1=(x^4+1)(x^4-1)}}} for starters.
{{{x^4+1=0}}} does not have real solutions, so you cannot factor {{{x^4+1}}}
However,
{{{x^4-1=(x^2)^2-1=(x^2+1)(x^2-1)}}}
{{{x^2+1=0}}} does not have real solutions, so you cannot factor {{{x^2+1}}}
However,
{{{x^2-1=(x+1)(x-1)}}}
Putting it all together, you get to the full factorization in a few steps
{{{x^8-1=(x^4)^2-1=(x^4+1)(x^4-1)=(x^4+1)((x^2)^2-1)=(x^4+1)(x^2+1)(x^2-1)=highlight((x^4+1)(x^2+1)(x+1)(x-1))}}}