$\displaystyle 1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1$

$\displaystyle P(1) = 1^3 = \frac{8}{8} = 1$

$\displaystyle P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2$

$\displaystyle P(k+1) = 1^3+...+k^3+(k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2$

from here I foiled and let m = p(k) for

$\displaystyle \left[ m + \frac{2(k+1)}{2}\right]^2$

$\displaystyle \left[ m+k+1\right]^2$