Answer
$\sum_{i=1}^{n}i^3 = [\frac{n(n+1)}{2}]^2$
Work Step by Step
$\sum_{i=1}^{n}(1+i)^4-i^4$
$= (2^4-1^4)+(3^4-2^4)+(4^4-3^4)+...+[(n+1)^4-n^4]$
$= (n+1)^4-1^4$
$= n^4+4n^3+6n^2+4n$
Also:
$\sum_{i=1}^{n}(1+i)^4-i^4$
$= \sum_{i=1}^{n}(4i^3+6i^2+4i+1)$
$= 4\sum_{i=1}^{n}i^3+ 6\sum_{i=1}^{n}i^2+ 4\sum_{i=1}^{n}i+ \sum_{i=1}^{n}1$
$= 4S+6(\frac{2n^2+3n^2+n}{6})+ 4(\frac{n^2+n}{2})+ n$
$= 4S+(2n^3+3n^2+n)+ (2n^2+2n)+ (n)$
$= 4S+2n^3+5n^2+4n$
We can equate these two expressions:
$4S+2n^3+5n^2+4n = n^4+4n^3+6n^2+4n$
$4S = n^4+2n^3+n^2$
$S = \frac{n^2(n^2+2n+1)}{4}$
$S = \frac{n^2(n+1)^2}{4}$
$S = [\frac{n(n+1)}{2}]^2$
Therefore:
$\sum_{i=1}^{n}i^3 = [\frac{n(n+1)}{2}]^2$
