Notice that
[tex](n+1)^4-n^4=4n^3+6n^2+4n+1[/tex]
so that
[tex]\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)[/tex]
We have
[tex]\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)[/tex]
[tex]\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1[/tex]
so that
[tex]\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)[/tex]
You might already know that
[tex]\displaystyle\sum_{i=1}^n1=n[/tex]
[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]
[tex]\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6[/tex]
so from these formulas we get
[tex]\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n[/tex]
[tex]\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4[/tex]
[tex]\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}[/tex]
If you don't know the formulas mentioned above:
