# How do you factor a^6+1?

Dec 9, 2015

Factor as a sum of cubes. Then we can factor the remaining quartic factor into two quadratics:

${a}^{6} + 1$

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

$= \left({a}^{2} + 1\right) \left({a}^{2} - \sqrt{3} a + 1\right) \left({a}^{2} + \sqrt{3} a + 1\right)$

#### Explanation:

The sum of cubes identity can be written:

${A}^{3} + {B}^{3} = \left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

So (putting $A = {a}^{2}$ and $B = 1$) we find:

${a}^{6} + 1 = {\left({a}^{2}\right)}^{3} + {1}^{3} = \left({a}^{2} + 1\right) \left({\left({a}^{2}\right)}^{2} - {a}^{2} + 1\right)$

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

It is not possible to factor $\left({a}^{2} + 1\right)$ into linear factors with Real coefficients, so let's leave that alone and look at the remaining quartic factor $\left({a}^{4} - {a}^{2} + 1\right)$. This does not factor as a quadratic in ${a}^{2}$ with Real coefficients, but will factor into two quadratics with irrational coefficients:

In order to get the terms in ${a}^{4}$, ${a}^{3}$, $a$ and the constant term to work out, the factorisation must be something like this:

${a}^{4} - {a}^{2} + 1 = \left({a}^{2} - k a + 1\right) \left({a}^{2} + k a + 1\right)$

Then the coefficient of ${a}^{2}$ is $2 - {k}^{2} = - 1$. Hence $k = \pm \sqrt{3}$.

${a}^{4} - {a}^{2} + 1 = \left({a}^{2} - \sqrt{3} a + 1\right) \left({a}^{2} + \sqrt{3} a + 1\right)$

Both of these quadratic factors has negative discriminants, so only Complex roots.

Dec 9, 2015

Alternatively, use Complex arithmetic to find the linear factors, then combine in conjugate pairs to derive the Real factoring:

${a}^{6} + 1 = \left({a}^{2} + 1\right) \left({a}^{2} - \sqrt{3} + 1\right) \left({a}^{2} + \sqrt{3} + 1\right)$

#### Explanation:

Using De Moivre's Theorem:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

Hence, the following Complex numbers are all $6$th roots of $- 1$:

$\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} i$

$\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = i$

$\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2} + \frac{1}{2} i$

$\cos \left(\frac{7 \pi}{6}\right) + i \sin \left(\frac{7 \pi}{6}\right) = - \frac{\sqrt{3}}{2} - \frac{1}{2} i$

$\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right) = - i$

$\cos \left(\frac{11 \pi}{2}\right) + i \sin \left(\frac{11 \pi}{2}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} i$

graph{(x^2+(y-1)^2-0.0015)(x^2+(y+1)^2-0.0015)((x-sqrt(3)/2)^2+(y-1/2)^2-0.0015)((x-sqrt(3)/2)^2+(y+1/2)^2-0.0015)((x+sqrt(3)/2)^2+(y+1/2)^2-0.0015)((x+sqrt(3)/2)^2+(y-1/2)^2-0.0015) = 0 [-2.5, 2.5, -1.25, 1.25]}

Taking these in conjugate pairs we find quadratic factors with Real coefficients:

$\left(a - i\right) \left(a + i\right) = {a}^{2} + 1$

$\left(a - \left(\frac{\sqrt{3}}{2} + \frac{1}{2} i\right)\right) \left(a - \left(\frac{\sqrt{3}}{2} - \frac{1}{2} i\right)\right) = {a}^{2} - \sqrt{3} a + 1$

$\left(a + \left(\frac{\sqrt{3}}{2} + \frac{1}{2} i\right)\right) \left(a + \left(\frac{\sqrt{3}}{2} - \frac{1}{2} i\right)\right) = {a}^{2} + \sqrt{3} a + 1$

Hence the factorisation with Real Coefficients is:

${a}^{6} + 1 = \left({a}^{2} + 1\right) \left({a}^{2} - \sqrt{3} + 1\right) \left({a}^{2} + \sqrt{3} + 1\right)$