Since $p$ is odd, we can express the left as the sum of $\frac{1}{k^3}+\frac{1}{(p-k)^3}$ where $1\le k\le\frac{p-1}{2}$. But $\frac{1}{k^3}+\frac{1}{(p-k)^3}=\frac{p((p-k)^2+k(p-k)+k^2)}{k^3(p-k)^3}$, so taking out $p$, the whole sum is $pa$ where $%Error. "latex" is a bad command. a$ is rational number whose denominator in the reduced state is not divisible by $%Error. "latex" is a bad command. p$. So $%Error. "latex" is a bad command. p|m$.