how to factorize the following function?

Question

I have calculate the following integral

$I=\int_{-\infty}^{\infty}\frac{x^{2}}{x^{6}+a^{6}}$, where $a>0$

I know by using residue theorem that:

$I=2\pi i\sum_{j}Res[f(z_{j})]$, for $Im(z)>0$

$f(z)=\frac{z^{2}}{z^{6}+a^{6}}$

in order to solve this I have to factorize f(z), but I don't know how to do it. I know that

$f(z)=\frac{z^{2}}{(z^{3}-a^{3}i)(z^{3}+a^{3}i)}$

this isn't enough to solve it, what do I do next?

I also tried:

$f(z)=\frac{z^{2}}{(z-a\sqrt[3]{i})(z^{2}+az\sqrt[3]{i}+a\sqrt[3]{i^{2}})(z+a\sqrt[3]{i})(z^{2}-az\sqrt[3]{i}+a\sqrt[3]{i^{2}})}$

with my second attempt I can use the formula for the residue:

$$Resf(z_{j})=\lim_{z\to z_{0}} (z-z_{0})f(z)$$

is this correct?

Answer 1

Hint:

Consider the fact that one root of $z^6+a^6$ is $ae^{\frac{i \pi}{6}}$ then integrate using the Residue Theorem the function $f(z)=\frac{z^{2}}{z^{6}+a^{6}}$ in the region delimited by:

$$\alpha_R=\{ z \in \mathbb{C}: Im(z)=0, 0 \leq |z| \leq R \}$$

$$ \beta_R=\{ Re^{it}: t \in [0,\theta] \}$$

$$ \gamma_R=\{z \in \mathbb{C}: z=e^{\frac{i \pi}{6}}t, t \in [0,R] \}$$

and the bring $R \to \infty$ getting the result.

Answer 2

You don't need complex analysis at all. Stay on the real line, do a bit of algebra, break up the expression into partial fractions and integrate. Here's how to factor the denominator. $$x^6+a^6=(x^2+a^2)(x^4-x^2a^2+a^4)$$ $$=(x^2+a^2)((x^2+a^2)^2-3a^2x^2)$$ $$=(x^2+a^2)((x^2+a^2)^2-(\sqrt3 ax)^2)$$ $$=(x^2+a^2)(x^2+a^2-\sqrt3 ax)(x^2+a^2+\sqrt3 ax)$$ $$=(x^2+a^2)(x^2-\sqrt3 ax+a^2)(x^2+\sqrt3 ax+a^2)$$ You can take it from here.