$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$ by Fourier series of $x^2$

We may start from $$ f_1(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n} \tag{1}$$ that is the Fourier series of a sawtooth wave, equal to $\frac{\pi-x}{2}$ on the interval $I=(0,2\pi)$.
By termwise integration, we get that $$ \forall x\in I,\quad \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2} = \frac{2\pi x-x^2}{4}$$ hence: $$ \forall x\in I,\quad f_2(x) = \sum_{n\geq 1}\frac{\cos(nx)}{n^2}=\frac{\pi^2}{6}-\frac{\pi x}{2}+\frac{x^2}{4}\tag{2} $$ $$ \forall x\in I,\quad f_3(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n^3}=\frac{\pi^2 x}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}\tag{3} $$ (the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem $$ \zeta(6) = \frac{1}{\pi}\int_{0}^{2\pi}f_3(x)^2\,dx = \frac{2\pi^6}{9}\int_{0}^{1}\left[x(x-1)(2x-1)\right]^2\,dx=\color{red}{\frac{\pi^6}{945}}.\tag{4}$$ With the same approach it is not difficult to prove that for any $n\geq 1$, $\zeta(2n)$ is a rational multiple of $\pi^{2n}\int_{0}^{1}B_n(x)^2\,dx$, where $B_n(x)$ is a Bernoulli polynomial.

Considering $f(x)=x^3$ By Parseval identity we can prove that $\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^6}{945}$. Then $$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=0,$$$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx=0$$ and \begin{align}b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\\&=(-1)^{n+1}\frac{2\pi^2}{n}+(-1)^{n}\frac{12\pi}{n^3}\end{align} From the relation $$\frac{1}{\pi}\int_{-\pi}^{\pi}|f|^2dx=\frac{a_0}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)$$we get \begin{align}\sum_{n=1}^\infty(\frac{144}{n^6}+\frac{4\pi^4}{n^2}-\frac{48\pi^2}{n^4})=\frac{2\pi^6}{7}\end{align}$$\sum_{n=1}^\infty\frac{144}{n^6}=\frac{16\pi^6}{105}$$ $$\sum_{n=1}^\infty\frac{1}{n^6}=\frac{\pi^6}{945}$$