calculus - Compute $sum_{n=1}^inftyfrac{H_n^3}{n^4}-3sum_{n=1}^inftyfrac{H_nH_n^{(2)}}{n^4}$

How to prove, without calculating each sum separately that

$$\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=24\zeta(7)-4\zeta(2)\zeta(5)-15\zeta(3)\zeta(4)\ ?$$
where $H_n^{(r)}=\sum_{k=1}^n\frac{1}{k^r}$ is the generalized harmonic number and $\zeta$ is Riemann zeta function.

---

This problem is already proved by Cornel in his book Almost Impossible Integrals, Sums and Series page 299 where he used only pure series manipulations.

The question here is can we prove it in a different way?

---

If you are curious about the result of each sum, you can find them in the same book page 301-302.

$$\sum_{n=1}^\infty\frac{H_n^3}{n^4}=\frac{231}{16}\zeta(7)+2\zeta(2)\zeta(5)-\frac{51}{4}\zeta(3)\zeta(4)$$

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=-\frac{51}{16}\zeta(7)+2\zeta(2)\zeta(5)+\frac{3}{4}\zeta(3)\zeta(4)$$