Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=1^3=1$ and $RHS=\frac{1}{4}(1^2)(1+1)^2=1$, thus it is true.
2. Assume the formula is true for $n=k$, we have
$1^3+2^3+3^3+...+k^3=\frac{1}{4}(k^2)(k+1)^2$
3. For $n=k+1$, we have $LHS=1^3+2^3+3^3+...+k^3+(k+1)^3=\frac{1}{4}(k^2)(k+1)^2+(k+1)^3=\frac{1}{4}(k+1)^2(k^2+4k+4)=\frac{1}{4}(k+1)^2(k+2)^2=RHS$
4. We conclude that the formula is true for any $n$.
