Compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$. 
Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.

I determined the coefficients of the Fourier series, which are
$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}$$
Then, I get
$$x^3 = \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(nx)$$
If $x = \pi$, then
$$\begin{aligned}
\pi^3 &= \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(n\pi)\\
\dfrac{3\pi^3}{4} &= \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}(-1)^n
\end{aligned}$$
I'm stuck.  It's easy to compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$, using the Fourier series, but for this type of problem I'm stuck.
Any comments or suggestions?  By the way, I know that
$$\sum\limits_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$
I need to know how to get there.

From your identity
\begin{equation*}
\frac{3\pi ^{3}}{4}=\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{
\pi n^{4}}(-1)^{n}
\end{equation*}
expanding the right hand side and using the result $\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi^2}{6}$, we get
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\sum_{n=1}^{\infty }\frac{6(\pi
^{2}n^{2}-2)(-1)^{n}+12}{n^{4}}(-1)^{n} \\
&=&6\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}-12\sum_{n=1}^{\infty }\frac{1
}{n^{4}}+12\sum_{n=1}^{\infty }\frac{(-1)^{n}}{n^{4}} \\
&=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-12\sum_{n=1}^{\infty }
\frac{(-1)^{n-1}}{n^{4}}.
\end{eqnarray*}
Now we need to express the alternating series $\sum_{n=1}^{\infty }\frac{
(-1)^{n-1}}{n^{4}}$ in terms of $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$, e.g.
as follows
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n-1}}{n^{4}} &=&\sum_{n=1}^{
\infty }\frac{1}{n^{4}}-2\sum_{n=1}^{\infty }\frac{1}{(2n)^{4}}
=\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{1}{2^{3}}\sum_{n=1}^{\infty }
\frac{1}{n^{4}}=\frac{7}{8}\sum_{n=1}^{\infty }\frac{1}{n^{4}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{
21}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}} \\
&=&\pi ^{4}-\frac{45}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}}.
\end{eqnarray*}
Solving for $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$ we finally obtain
\begin{equation*}
\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{2}{45}\left( \pi ^{4}-\frac{3\pi
^{4}}{4}\right) =\frac{\pi ^{4}}{90}.
\end{equation*}
