How is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$

If you already know or take as given the first result, then

$$\frac{\pi^4}{90}=\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}\implies$$

$$\sum_{n=1}^\infty\frac1{(2n-1)^4}=\left(1-\frac1{16}\right)\frac{\pi^4}{90}=\frac{\pi^4}{96}$$