I've done the first part of the question but I'm not sure about the second part. I have some special results that might be required for it but they possibly aren't required.
$\displaystyle \displaystyle 1^3 +2^3+3^3+ ... + n^3 = \frac {n^2}{4} (n+1)^2 $
$\displaystyle n=1 $
$\displaystyle \displaystyle 1^3 = \frac {1^2}{4} ( 1+1) ^2 \Rightarrow 1=1 $
(show for n=2, n=3 )
assume that it's true for n=k :
$\displaystyle \displaystyle 1^3 +2^3+3^3+ ... + k^3 = \frac {k^2}{4} (k+1)^2 $
now with assumption that is true for n=k , show that is true for n=k+1 ....
$\displaystyle \displaystyle 1^3 +2^3+3^3+ ... + k^3 +(k+1)^3= \frac {(k+1)^2}{4} ((k+1)+1)^2 $
can you do that ?
hint : $\displaystyle \displaystyle 1^3 +2^3+3^3+ ... + k^3 = \frac {k^2}{4} (k+1)^2 $ so add $\displaystyle \displaystyle(k+1)^3$ to $\displaystyle \frac {k^2}{4} (k+1)^2$ and you should get the $\displaystyle \displaystyle \frac {(k+1)^2}{4} ((k+1)+1)^2 $
hmmm.... okay
did you try to do it with that identity ?
$\displaystyle \displaystyle (1\cdot 2 \cdot 3 - 3\cdot 1 \cdot 2 +1 )^3 + ( 2\cdot 3 \cdot 4 - 3\cdot 2 \cdot 3 +1)^3 + ... + (n\cdot (n+1) \cdot (n+2) - 3n \cdot (n+1) +n)^3 = \frac {n^2}{4} \cdot (n+1)^2 $
and do induction