Combinatorial proof of the identity: $$\sum_{k=1}^n {(k^3)} = \left(\sum_{k=1}^n {k}\right)^2 $$


How could I approach this problem to make a combinatorial proof?
Try induction.  Suppose the statement holds for $n-1$.  To prove it for $n$, write:
$$
\left( \sum_{k=1}^{n} k \right)^2 = \left( n + \sum_{k=1}^{n-1} k \right)^2 = n^2 + 2 n \sum_{k=1}^{n-1} k + \left( \sum_{k=1}^{n-1} k \right)^2
= 
n^2 + 2 n \sum_{k=1}^{n-1} k +
\sum_{k=1}^{n-1} k^3.
$$
Now use
$$
\sum_{k=1}^{n-1} k = {(n-1) \, n \over 2}.
$$
Try induction.  Suppose the statement holds for $n-1$.  To prove it for $n$, write:
$$
\left( \sum_{k=1}^{n} k \right)^2 = \left( n + \sum_{k=1}^{n-1} k \right)^2 = n^2 + 2 n \sum_{k=1}^{n-1} k + \left( \sum_{k=1}^{n-1} k \right)^2
= 
n^2 + 2 n \sum_{k=1}^{n-1} k +
\sum_{k=1}^{n-1} k^3.
$$
Now use
$$
\sum_{k=1}^{n-1} k = {(n-1) \, n \over 2}.
$$