1. Find the value of the expression $(3 x-y)(x+3 y)+(3 x+y)(x-3 y)$ given that $x^{2}+y^{2}+4 x-10 y+29=0$.
Solution. We have

$$
x^{2}+y^{2}+4 x-10 y+29=(x+2)^{2}+(y-5)^{2}
$$

it follows that

$$
(x+2)^{2}+(y-5)^{2}=0
$$

From this, $x+2=0$ and $y-5=0$, hence $x=-2$ and $y=5$. Therefore,

$$
(3 x-y)(x+3 y)+(3 x+y)(x-3 y)=-11 \cdot 13+17=-126
$$