4 . Find the sum of $1^{3}+2^{3}+3^{3}+\cdots+1984^{3}$.
Solving: Using the proof from the previous question that the cube of any natural number can be expressed as the difference of the squares of two natural numbers, we can find the sum of the cubes of the first $n$ natural numbers:
$$
1^{3}+2^{3}+3^{3}+\cdots+n^{3} \text {. }
$$

Substituting $1,2,3 \cdots, n$ into the identity (1) from the previous question, we get:
$$
\begin{array}{l}
1^{3}=1^{2}-0^{2}, \\
2^{3}=3^{2}-1^{2}, \\
3^{3}=6^{2}-3^{2}, \\
\cdots \cdots \\
n^{3}=\left[\frac{n(n+1)}{2}\right]^{2}-\left[\frac{n(n-1)}{2}\right]^{2} .
\end{array}
$$

Adding the left and right sides of the above $n$ equations, we obtain:
$$
1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2} \text {. }
$$

When $n=1984$, substituting into identity (2), we get:
$$
1^{3}+2^{3}+3^{3}+\cdots+1984^{3}=\left[\frac{1984 \cdot 1985}{2}\right]^{2} .
$$