Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=1^3=1$, $RHS=1^2(2-1)=1$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $1^3+3^3+5^3+...+(2k-1)^3=k^2(2k^2-1)$
Step 3. Prove it is true for $n=k+1$: $LHS=1^3+3^3+5^3+...+(2k-1)^3+(2k+1)^3=k^2(2k^2-1)+(2k+1)^3=2k^4-k^2+8k^3+12k^2+6k+1=2k^4+8k^3+11k^2+6k+1$
Use $n=k+1$ on the right side of the formula, we have $RHS=(k+1)^2(2(k+1)^2-1)=(k^2+2k+1)(2k^2+4k+1)
=2k^4+4k^3+k^2+4k^3+8k^2+2k+2k^2+4k+1=2k^4+8k^3+11k^2+6k+1$
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.
