Prove that $\sum_{t \vert n} d^3(t) = (\sum_{t \vert n}d(t))^2$ for all $n \in \mathbb{N}$
Prove that if $n \in \mathbb{N}$, then $\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$ where $d(n)$ is the divisor function.