Evaluate the sum of $1/n^6$ using Euler's method

Question

I've just learned how Euler evaluated $1+1/2^2+1/3^2+...+1/n^2+...=\pi^2/6$ by comparing the coefficients of the series form and product form of $\sin(x)/x$.

The series form is $$\dfrac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots \tag{1}$$ The product form is $$\frac{\sin(x)}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\ldots\left(1-\frac{x^2}{n^2\pi^2}\right)\ldots\tag{2}$$

To evaluate $1+1/2^4+1/3^4+...+1/n^4+...=\pi^4/90$, we can multiply $\sin(x)/x$ by $\sin(\mathrm{i}x)/(\mathrm{i}x)$.

The series form is $$\dfrac{\sin(x)}{x}\dfrac{\sin(\mathrm{i}x)}{\mathrm{i}x}=\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots\right)\left(1-\frac{(\mathrm{i}x)^2}{3!}+\frac{(\mathrm{i}x)^4}{5!}-\ldots\right) \tag{3}$$ The product form is $$\frac{\sin(x)}{x}\dfrac{\sin(\mathrm{i}x)}{\mathrm{i}x}=\left(1-\frac{x^4}{\pi^4}\right)\left(1-\frac{x^4}{2^4\pi^4}\right)\ldots\left(1-\frac{x^4}{n^4\pi^4}\right)\ldots\tag{4}$$

Compare the coefficient of $x^4$ and we can get the result of $\pi^4/90$.

Now I want to evaluate $\sum 1/n^6$. How can it be solved? $(1-x^6/\pi^6)$ can be obtained from $(1-x^3/\pi^3)$. But we cannot find $(1-x^3/\pi^3)$. It can't be obtained from $(1-x^4/\pi^4)$ either. This really confuses me. How can I obtain the wanted form for comparison?

Thank you in advance!

Answer 1

This is another way to calculate the sum of $4$th powers, the source I used said that the method for the $6$th powers is very similar: $$\frac{\pi^4}{36}= \sum_{i=1}^{\infty}\frac{1}{i^2}=\sum_{i,j\in{[1,\infty]\cap{\natural}}}\frac{1}{i^2j^2}=\sum_{i<j}\frac{1}{i^2j^2}+\sum_{i>j}\frac{1}{i^2j^2}+\sum_{i=j}\frac{1}{i^2j^2}=2\sum_{i>j}\frac{1}{i^2j^2} +\sum_{i=1}^{\infty}\frac{1}{i^4}$$ The first sum is the $x^5$ successor in the product times $\pi^4$, and the second is what we are after source (if my explanation is poor) : https://youtu.be/WL_Yzbo1ha4?si=yw8PVGLCTt4Oq5U_