## Task $4 / 86$

Determine a method by which any cube number can be represented as the difference of the squares of two natural numbers, and derive from this a formula for the sum of the first $n$ cube numbers!
It should hold that $k^{3}=a_{k}^{2}-b_{k}^{2}$ with $k ; a_{k} ; b_{k} \in \mathbb{N}$, thus $k^{2} \cdot k=\left(a_{k}+b_{k}\right)\left(a_{k}-b_{k}\right)$. Therefore, we can set $k^{2}=a_{k}+b_{k}, k=a_{k}-b_{k}$. From this, it follows that

$$
a_{k}=\frac{k^{2}+k}{2}=\frac{k(k+1)}{2} \quad ; \quad b_{k}=\frac{k^{2}-k}{2}=\frac{k(k-1)}{2}
$$

Since the product of any two consecutive natural numbers is even, it follows that $a_{k}$ and $b_{k}$ are integers. For every $k$, it holds that $b_{k+1}=\frac{(k+1) k}{2}=a_{k}$. From this, it follows that

$$
\begin{gathered}
\sum_{k=1}^{n} k^{3}=\sum_{k=1}^{n}\left(a_{k}^{2}-b_{k}^{2}\right)=\sum_{k=1}^{n} a_{k}^{2}-\sum_{k=1}^{n} b_{k}^{2}=a_{n}^{2}+\sum_{k=1}^{n-1} a_{k}^{2}-\sum_{k=2}^{n} b_{k}^{2}-b_{1}^{2}= \\
=a_{n}^{2}+\sum_{k=1}^{n-1} a_{k}^{2}-\sum_{k=1}^{n-1} b_{k}^{2}+0=a_{n}^{2}+\sum_{k=1}^{n-1} a_{k}^{2}-\sum_{k=1}^{n-1} a_{k}^{2}=a_{n}^{2}=\left(\frac{n(n+1)}{2}\right)^{2}=\frac{n^{2}(n+1)^{2}}{4}
\end{gathered}
$$