FInd $\displaystyle \sum^{n}_{r=1}r^3$

My attempt :

Using this identity : $\displaystyle r^4-(r-1)^4=4r^3-6r^2+4r-1$

$\displaystyle \sum^{n}_{r=1}r^4-(r-1)^4=\sum^{n}_{r=1}4r^3-6r^2+4r-1$

The LHS can be written as $\displaystyle n^4$ .

$\displaystyle 4\sum^{n}_{r=1}r^3=n^4+n(n+1)(2n+1)+2n(n+1)-n$

but it is not anywhere nearer to $\displaystyle \frac{1}{4}n^2(n+1)^2 $ ,there must be some problem with my proving .