$$
m^3=n^3+n^2p
$$

let $m=n+k$ , then
$$
3n^2k+3nk^2+k^3=n^2p\\
(p-3k)n^2-3k^2n-k^3=0
$$
witch $p>3k$ . solving this function, we get
$$
n = \frac{3k^2 \pm k\sqrt{4kp-3k^2}}{p-3k}
$$