Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=1^3=1$, $RHS=\frac{1^2(1+1)^2}{4}=1$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $1^3+2^3+3^3+...+k^3=\frac{k^2(k+1)^2}{4}$
Step 3. Prove it is true for $n=k+1$: $LHS=1^3+2^3+3^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k^2+4k+4)}{4}=\frac{(k+1)^2(k+2)^2}{4}$
Use $n=k+1$ on the right side of the formula, we have $RHS=\frac{(k+1)^2(k+2)^2}{4}$
Thus, we have $LHS=RHS$ which proves the formula is also true for $n=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the formula is true for all natural numbers n.
