Prove $1^{3} + 2^{3} + ... + n^{3}$ = $(1 + 2 + ... + n)^{2}$ for all positive integers n.
I've tried to work with this problem using mathematical induction. However, I really don't understand how to manipulate the right side of the equation. Any help would be greatly appreciated.
Thank you in advance.
It works for $n=1$. $$\begin{align} 1^3+\cdots+n^3+(n+1)^3 & =(1+\cdots +n)^2+(n+1)^3 \quad \mathrm{induction} \ \mathrm{hypothesis} \\ &= \frac{n^2(n+1)^2+4(n+1)^3}{4} \\ &=\frac {(n+1)^2(n^2+4(n+1))}{4} \\ &=\frac {(n+1)^2((n+1)+1)^2}{4} \\ &=(1+\cdots +n+(n+1))^2\end{align}$$
Assume the result holds for $n$. Then
$$1^3+2^3+\dotsb+n^3+(n+1)^3 = (1+2+\dotsb+n)^2+(n+1)^3$$
$$(1+2+\dotsb+n+(n+1))^2 = (1+2+\dotsb+n)^2 + 2(1+2+\dotsb+n)(n+1)+(n+1)^2 \\=1^3+2^3+\dotsb + n^3+2\left(\frac{n(n+1)}{2}\right)(n+1)+(n+1)^2 \\=1^3+2^3+\dotsb + n^3+n(n+1)(n+1)+(n+1)^2=1^3+2^3+\dotsb + n^3+(n+1)^3$$
