Your idea matters-20(Reuploaded)

For every single integer $n$ from $1$ to $\infty$ , $1^3+2^3+3^3+\cdots+n^3$ will equal $(1+2+3+\cdots+n)^2$ .

Proof:

Each of the squares has a different integer length, shown at the side of the graph. If you notice, we have one square with an area of $1^2$ , two square with an area of $2^2$ , three squares with an area of $3^3$ , etc. In other words, $1(1^2)+2(2^2)+3(3^2)+\cdots+n(n^2)=1^3+2^3+3^3+\cdots+n^3$ .

If you also notice, each time, the squares come to together to make a larger square (more or less). That square is actually equal to $(1+2+3+\cdots+n)^2$ . However, in some of the cases, the squares overlap and leave a hole. However, the overlap and the hole are exactly equal, so we can say that $(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3$ .