I assume the claim is being made for integers [tex]n\ge1[/tex], not just any integer.
Basis step: Let [tex]n=1[/tex]. Then
[tex]\displaystyle \sum_{i=1}^1 i^3 = 1^3 = 1[/tex]
[tex]\dfrac{1^2(1+1)^2}4 = \dfrac{2^2}4 = 1[/tex]
so this case holds.
Inductive step: Assume
[tex]\displaystyle \sum_{i=1}^k i^3 = \frac{k^2 (k+1)^2}4[/tex]
As a result, we want to show that
[tex]\displaystyle \sum_{i=1}^{k+1} i^3 = \frac{(k+1)^2 (k+2)^2}4[/tex]
Now,
[tex]\displaystyle \sum_{i=1}^{k+1} i^3 = \sum_{i=1}^k i^3 + (k+1)^3 \\\\ ~~~~~~~~ = \frac{k^2(k+1)^2}4 + (k+1)^3 \\\\ ~~~~~~~~ = \frac{k^2(k+1)^2 + 4(k+1)^3}4 \\\\ ~~~~~~~~ = \frac{(k+1)^2 \bigg(k^2 + 4(k+1)\bigg)}4 \\\\ ~~~~~~~~ = \frac{(k+1)^2 (k^2 + 4k + 4)}4 \\\\ ~~~~~~~~ = \frac{(k+1)^2 (k+2)^2}4[/tex]
QED
