Answer
See below.
Work Step by Step
1. For $n=1$, we have $LHS=1^3=1$ and $RHS=[\frac{1(1+1)}{2}]^2=1=LHS$, thus P(1) is true.
2. Assume P(k) ($k\gt1$) is true, we have:
$1^3+2^3+...+k^3=[\frac{k(k+1)}{2}]^2$
3. For $n=k+1$, we have:
$LHS=1^3+2^3+...+k^3+(k+1)^{3}=[\frac{k(k+1)}{2}]^2+(k+1)^{3}\\
=\frac{k^2(k+1)^2+4(k+1)^3}{4}=\frac{(k+1)^2(k^2+4k+4)}{4}\\
=\frac{(k+1)^2(k+2)^2}{4}=[\frac{(k+1)(k+2)}{2}]^2=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.