Solution:

We can factor
$$p = a^{3} - b^{3} = (a-b)\left(a^{2} + a b + b^{2}\right).$$
Since $a$ and $b$ are positive integers, the only way this can happen is if $a-b=1$.
Either $a$ or $b$ is even. If $a$ is even, let $a=2u$, so $b=2u-1$. Then
$$\begin{aligned} p & = (2u)^{2} + (2u)(2u-1) + (2u-1)^{2} \\ & = 12u^{2} - 6u + 1 \\ & = (3u-1)^{2} + 3u^{2} \end{aligned}$$
has the desired form. If $b$ is even, let $b=2u$, so $a=2u+1$. Then
$$\begin{aligned} p & = (2u+1)^{2} + (2u)(2u+1) + (2u)^{2} \\ & = 12u^{2} + 6u + 1 \\ & = (3u+1)^{2} + 3u^{2} \end{aligned}$$
has the desired form.