# How do you factor (x^6)+1?

May 10, 2015

${x}^{6} + 1 = \left({x}^{4} - {x}^{2} + 1\right) \left({x}^{2} + 1\right)$

How did I find this? First of all, substitute ${x}^{2} = y$ to get ${y}^{3} + 1$.

Notice that $y = - 1$ is a solution of ${y}^{3} + 1 = 0$, so $\left(y + 1\right)$ is a factor.

We can easily derive ${y}^{3} + 1 = \left({y}^{2} - y + 1\right) \left(y + 1\right)$, noticing the way that the intermediate terms cancel out when you multiply this out.

Substituting $y = {x}^{2}$ back in, we get:
${x}^{6} + 1 = \left({x}^{4} - {x}^{2} + 1\right) \left({x}^{2} + 1\right)$

How do we know this factorization is complete?

There is no linear factor, because ${x}^{6} + 1$ is always > 0 (for real values of $x$).

Suppose ${x}^{4} - {x}^{2} + 1 = \left({x}^{2} + a x + b\right) \left({x}^{2} + c x + d\right)$.
Looking at the coefficient of ${x}^{3}$, we get $a + c = 0$, so $c = - a$:
... $= \left({x}^{2} + a x + b\right) \left({x}^{2} - a x + d\right)$
The constant term gives us that $b d = 1$ so $b = d = 1$ or $b = d = - 1$.
The coefficient of ${x}^{2}$ gives us ${a}^{2} - \left(b + d\right) = 1$. Hence ${a}^{2} = 3$ or ${a}^{2} = - 1$. Neither have integer solutions.