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$\sum_{n=1}^{\infty}\sum_{j=1}^{n}\sum_{k=1}^{\infty}\dfrac{1}{j\left(n+k\right)^3}$ $=\sum_{j=1}^{\infty}\dfrac{1}{j}\sum_{n=j}^{\infty}\sum_{k=1}^{\infty}\dfrac{1}{\left(n+k\right)^3}$ $=\sum_{j=1}^{\infty}\dfrac{1}{j}\sum_{i=1}^{\infty}\dfrac{i}{\left(j+i\right)^3}$ $=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\dfrac{1}{\left(j+i\right)^3}\dfrac{i}{j}$ Summing across all $p=j+i$ gives $\sum_{p=2}^{\infty}\dfrac{1}{p^3}\sum_{q=1}^{p-1}\dfrac{p-q}{q}$ $=\sum_{p=2}^{\infty}\dfrac{1}{p^3}\left(pH_{p-1}-p+1\right)$ $=\sum_{p=2}^{\infty}\dfrac{H_{p-1}}{p^2}-\dfrac{1}{p^2}+\dfrac{1}{p^3}$ $=\sum_{p=1}^{\infty}\dfrac{H_{p}}{p^2}-\dfrac{1}{p^2}$ It is known that $\sum_{p=1}^{\infty}\dfrac{H_{p}}{p^2}=2\zeta\left(3\right)$ . Hence $\sum_{n=1}^{\infty}\sum_{j=1}^{n}\sum_{k=1}^{\infty}\dfrac{1}{j\left(n+k\right)^3}=\boxed{2\zeta\left(3\right)-\zeta\left(2\right)}$