Need help on the proof of $\displaystyle \sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)$

The series of the squares of the first n natural numbers , ie

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

I know this identity is needed to complete the proof :

$\displaystyle

(r+1)^3=r^3+3r^3+3r+1

$

From here onwards , i am not sure what to do ..

Not sure how to proof this as well :

$\displaystyle

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

$