x8+x4+1
2 answers
12x+1
There are clearly no real solutions, but you can factor it into
(x^2-x+1)(x^2+x+1)(x^4-x^2+1) = 0
because
x^8+x^4+1 = x^8+2x^4+1 - x^4
= (x^4+1)^2 - (x^2)^2
a difference of two squares
Now you can find the roots of those quadratics
(x^2-x+1)(x^2+x+1)(x^4-x^2+1) = 0
because
x^8+x^4+1 = x^8+2x^4+1 - x^4
= (x^4+1)^2 - (x^2)^2
a difference of two squares
Now you can find the roots of those quadratics