Find all triples $(a, b, c)$ of non-negative integers satisfying $a \geqslant b \geqslant c$ and $1 \cdot a^{3}+9 \cdot b^{2}+9 \cdot c+7=1997$.
Answer: $(10,10,10)$ is the only such triple.

The equality immediately implies $a^{3}+9 b^{2}+9 c=1990 \equiv 1(\bmod 9)$. Hence $a^{3} \equiv 1(\bmod 9)$ and $a \equiv 1(\bmod 3)$. Since $13^{3}=2197>1990$ then the possible values for $a$ are $1,4,7,10$.

On the other hand, if $a \leqslant 7$ then by $a \geqslant b \geqslant c$ we have

$$
a^{3}+9 b^{2}+9 c^{2} \leqslant 7^{3}+9 \cdot 7^{2}+9 \cdot 7=847<1990
$$

a contradiction. Hence $a=10$ and $9 b^{2}+9 c=990$, whence by $c \leqslant b \leqslant 10$ we have $c=b=10$.