# How do you factor x^6-26x^3-27?

Feb 5, 2015

First let's notice that ${x}^{6} = {\left({x}^{3}\right)}^{2}$

Let $r = {x}^{3}$

Using substitution we have

${r}^{2} - 26 r - 27$

We now look for the factors of $- 27$ that will combine (add) to make $- 26$

We find that $- 27$ and $1$ multiply to be $- 27$ and add to be $- 26$. This gives us

$\left(r - 27\right) \left(r + 1\right)$

Now substitute ${x}^{3}$ back in for $r$

$\left({x}^{3} - 27\right) \left({x}^{3} + 1\right)$

We should recognize this as a difference of cubes and a sum of cubes.

$\left(x - 3\right) \left({x}^{2} + 3 x + 9\right) \left(x + 1\right) \left({x}^{2} - x + 1\right)$