Prove the sum of cubes formula:
$$1^{3}+2^{3}+\cdots+(n-1)^{3}+n^{3}=\frac{n^{2}(n+1)^{2}}{4} \text {. }$$

In fact, it is only necessary to prove that $97^{97}$ cannot be expressed as the sum of cubes of several consecutive positive integers.
Assume there exist $m, n \in \mathbf{N}$, such that
$$97^{97}=(m+1)^{3}+(m+2)^{3}+\cdots+(n-1)^{3}+n^{3} \text {. }$$

Then $97^{97}=\left(1^{3}+2^{3}+\cdots+n^{3}\right)-$
$$\begin{array}{r} \left(1^{3}+2^{3}+\cdots+m^{3}\right) \\ =\frac{n^{2}(n+1)^{2}}{4}-\frac{m^{2}(m+1)^{2}}{4} . \end{array}$$

Thus, $n^{2}(n+1)^{2}-m^{2}(m+1)^{2}=4 \times 97^{97}$.
Using the difference of squares formula, factorize to get
$$\begin{array}{l} (n(n+1)+m(m+1))(n(n+1)-m(m+1)) \\ =4 \times 97^{97} . \end{array}$$

Set $n(n+1)+m(m+1)=2 \times 97^{s}\left(s \in \mathbf{N}_{+}\right)$, $n(n+1)-m(m+1)=2 \times 97^{t}(t \in \mathbf{N})$.
Then $n(n+1)-m(m+1)$
$$=(n+m+1)(n-m)=2 \times 97^{t} \text {. }$$

Consider two cases.
(1) $\left\{\begin{array}{l}n+m+1=2 \times 97^{r} \\ n-m=97^{t-r}\end{array},\left(r \in \mathbf{N}_{+}, t \geqslant r\right)\right.$.

Solving, we get $\left\{\begin{array}{l}m=\frac{2 \times 97^{r}-97^{t-r}-1}{2}, \\ n=\frac{2 \times 97^{r}+97^{t-r}-1}{2} .\end{array}\right.$
Thus, $n(n+1)+m(m+1)$
$=\frac{\left(2 \times 97^{r}+97^{t-r}\right)^{2}+\left(2 \times 97^{r}-97^{t-r}\right)^{2}-2}{4}$
$=97^{2 r}+\frac{97^{2 t-2 r}-1}{2}=2 \times 97^{s}$.
Therefore, $2 \times 97^{2 r}+97^{2 t-2 r}-1=4 \times 97^{s}$.
Since $s \geqslant 1, r \geqslant 1$, we have $2 t-2 r=0$ (otherwise, the left side of equation (1) cannot be divisible by 97).
Thus, $97^{2 r}=2 \times 97^{s}$, which is impossible.
(2) $\left\{\begin{array}{l}n+m+1=97^{r}, \\ n-m=2 \times 97^{t-r}\end{array}\left(r \in \mathbf{N}_{+}, t \geqslant r\right)\right.$.
Solving, we get $\left\{\begin{array}{l}m=\frac{97^{r}-2 \times 97^{t-r}-1}{2}, \\ n=\frac{97^{r}+2 \times 97^{t-r}-1}{2} .\end{array}\right.$
Thus, $n(n+1)+m(m+1)$
$$\begin{array}{l} =\frac{\left(97^{r}+2 \times 97^{t-r}\right)^{2}+\left(97^{r}-2 \times 97^{t-r}\right)^{2}-2}{4} \\ =\frac{97^{2 r}-1}{2}+97^{2 t-2 r}=2 \times 97^{s} . \end{array}$$

Therefore, $97^{2 r}-1+2 \times 97^{2 t-2 r}=4 \times 97^{s}$.
Since $s \geqslant 1, r \geqslant 1$, we have $2 t-2 r=0$ (otherwise, the left side of equation (2) cannot be divisible by 97).
Thus, $97^{2 r}+1=2 \times 97^{s}$, which is impossible.
In conclusion, $97^{97}$ cannot be expressed as the sum of cubes of several consecutive positive integers.