Prove that The sum of the cube of the first n odd numbers = $n^2(2n^2-1)$

That is

$$\sum_{i=1}^n (2i-1)^3 = n^2(2n^2-1)$$

First I tried to solve it by induction:

$Basis-step:$

For $n=1$ we have

$$\sum_{i=1}^n (2i-1)^3 = 1$$

and $$n^2(2n^2-1) = 1$$

$Induction-step:$

Let $n \in \mathbb{N},$ $$Assume \sum_{i=1}^n (2i-1)^3 = n^2(2n^2-1)$$ $$Prove \sum_{i=1}^{n+1} (2i-1)^3 = (n+1)^2(2(n+1)^2-1)$$

I know that $$\sum_{i=1}^{n+1} (2i-1)^3 = \sum_{i=1}^n (2i-1)^3 + (2(n+1)-1)^3 $$ But i was not able to reach a solution.

If anyone can give a small hint, that would be helpful.

## Best Answer

All that's missing is to prove that$$\bigl(2(n+1)-1\bigr)^3+n^2(2n^2-1)=(n+1)^2\bigl(2(n+1)^2-1\bigr).$$But if you expand both sides of this equality, you will get $2 n^4+8 n^3+11 n^2+6 n+1$ in each case.