Problem 10.8. Real numbers $x$ and $y$ are such that $x^{3}+21 x y+y^{3}=343$. What can $x+y$ be? List all possible options.
Answer: $7, -14$.

Solution. By directly expanding the brackets, it is easy to verify the identity

$$
a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)
$$

This means that the expression on the left side can be zero either when $a+b+c=0$,
or when $a^{2}+b^{2}+c^{2}-a b-b c-c a=0$, which is only possible when $a=b=c$, since

$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}
$$

The condition of the problem is equivalent to the equality

$$
x^{3}+y^{3}+(-7)^{3}-3 x y \cdot(-7)=0
$$

From the above, it follows that this is true either when $x+y-7=0$ (i.e., $x+y=7$), or when $x=y=-7$ (in which case $x+y=-14$).