# How to factor a^8+b^8 ?

Oct 10, 2016

${a}^{8} + {b}^{8} = {\prod}_{k = 0}^{7} \left(a - | b | {e}^{i \pi \frac{2 k + 1}{8}}\right)$ for $b \in \mathbb{R}$

${a}^{8} + {b}^{8} = {\prod}_{k = 0}^{7} \left(a - | b | {e}^{i \pi \left(\frac{\theta}{\pi} + \frac{2 k + 1}{8}\right)}\right)$ for $b = | b | {e}^{i \theta} \in \mathbb{C}$

#### Explanation:

By the fundamental theorem of algebra, we can factor the given expression as

${a}^{8} + {b}^{8} = {\prod}_{k = 1}^{8} \left(a - {\alpha}_{k}\right)$

where each ${\alpha}_{k}$ is a root of ${x}^{8} + {b}^{8}$.

Solving for ${\alpha}_{k}$, we get

${x}^{8} + {b}^{8} = 0$

$\implies {x}^{8} = - {b}^{8}$

$\implies x = {\left(- {b}^{8}\right)}^{\frac{1}{8}}$

$= | b | {\left(- 1\right)}^{\frac{1}{8}}$ (assuming $b \in \mathbb{R}$)

$= | b | {\left({e}^{i \left(\pi + 2 \pi k\right)}\right)}^{\frac{1}{8}}$

=|b|e^(ipi((2k+1)/8), k in ZZ

As $k \in \left\{0 , 1 , 2 , 3 , 4 , 5 , 6 , 7\right\}$ accounts of all unique values of that form, we get our factorization as, for $b \in \mathbb{R}$

${a}^{8} + {b}^{8} = {\prod}_{k = 0}^{7} \left(a - | b | {e}^{i \pi \frac{2 k + 1}{8}}\right)$

For a more general $b \in \mathbb{C}$, then supposing $b = | b | {e}^{i \theta}$, we can go through similar calculations to find

${\left(- {b}^{8}\right)}^{\frac{1}{8}} = | b | {e}^{i \pi \left(\frac{\theta}{\pi} + \frac{2 k + 1}{8}\right)}$

meaning

${a}^{8} + {b}^{8} = {\prod}_{k = 0}^{7} \left(a - | b | {e}^{i \pi \left(\frac{\theta}{\pi} + \frac{2 k + 1}{8}\right)}\right)$

Oct 10, 2016

Sorry, I overlook some minor details, the answer provided by sente is correct.

Oct 11, 2016

Supposing $b \ne 0$ and $a , b \in \mathbb{R}$ we have
${\left(\frac{a}{b}\right)}^{8} = - 1 = {e}^{i \pi + 2 k \pi}$ then
$\frac{a}{b} = {e}^{i \left(2 k + 1\right) \frac{\pi}{8}}$ then
$a - b {e}^{i \left(2 k + 1\right) \frac{\pi}{8}} = 0$ are the $k = 0 , 1 , \cdots , 7$ roots or factors.

Define

$p \left(k\right) = a - b {e}^{i \left(2 k + 1\right) \frac{\pi}{8}}$

and then

${f}_{1} = p \left(1\right) p \left(6\right) = {a}^{2} - \left(\sqrt{2 - \sqrt{2}}\right) a b + {b}^{2}$
${f}_{2} = p \left(2\right) p \left(5\right) = {a}^{2} + \left(\sqrt{2 - \sqrt{2}}\right) a b + {b}^{2}$
${f}_{3} = p \left(3\right) p \left(4\right) = {a}^{2} + \left(\sqrt{2 + \sqrt{2}}\right) a b + {b}^{2}$
${f}_{4} = p \left(0\right) p \left(7\right) = {a}^{2} - \left(\sqrt{2 + \sqrt{2}}\right) a b + {b}^{2}$

so

${a}^{8} + {b}^{8} = {f}_{1} {f}_{2} {f}_{3} {f}_{4}$ with real coefficients.