Question 635918
To factor {{{x^6+1}}} , I would think of
{{{y^3+1=(y+1)(y^2-y+1)}}} , which is a full factorization,
because if {{{y^2-y+1}}} had factors (with real coefficients),
{{{y^2-y+1=0}}} would have real solutions (and it does not).
If we make {{{y=x^2}}} in {{{y^3+1=(y+1)(y^2-y+1)}}} , we get
{{{highlight(x^6+1=(x^2+1)(x^4-x^2+1))}}}