calculus - How prove this $sum_{n=1}^{infty}frac{zeta_{2}}{n^4}=zeta^2(3)-frac{1}{3}zeta(6)$

show that
$$\sum_{n=1}^{\infty}\dfrac{\zeta_{2}}{n^4}=\zeta^2(3)-\dfrac{1}{3}\zeta(6)$$

where
$$\zeta_{m}=\sum_{k=1}^{n}\dfrac{1}{k^m},\zeta(m)=\sum_{k=1}^{\infty}\dfrac{1}{k^m}$$
is true？
because This result is my frend tell me.

This problem have someone research it?Thank you

my some idea:
$$\zeta^3(3)=\left(\sum_{n=0}^{\infty}\dfrac{1}{(n+1)^3}\right)^2=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\dfrac{1}{(k+1)^3(n-k+1)^3}$$

and use
$$\dfrac{1}{(k+1)(n-k+1)}=\dfrac{1}{n+2}\left(\dfrac{1}{k+1}+\dfrac{1}{n-k+1}\right)$$
and $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
and
$$\sum_{n=1}^{\infty}\dfrac{H_{n}}{(n+1)^5}=\dfrac{1}{2}\left(5\zeta(6)-2\zeta(2)\zeta(4)-\zeta^2(3)\right)$$

But is very ugly, someone have other nice methods? Thank you .