Solve the following equation:

$$
2 x^{4}+2 y^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+7 y^{2}+7 z^{2}-14 y z-70 y+70 z+175=0 .
$$
Solution. The left side of the equation can be written as a sum of powers with even exponents:

$$
\begin{aligned}
(x-y)^{4}+x^{4}+y^{4}+7\left(y^{2}+z^{2}-2 y z-10 y+10 z+25\right) & =0 \\
(x-y)^{4}+x^{4}+y^{4}+7(z-y+5)^{2} & =0
\end{aligned}
$$

The value of a power with an even exponent cannot be negative, so the sum can only be 0 if all terms are simultaneously equal to zero.

For this, the following conditions must be satisfied simultaneously:

$$
\begin{array}{rlrlrrrr}
(x-y)^{4}=0 & \text { and } & x^{4}=0 & \text { and } & y^{4}=0 & \text { and } & 7(z-y+5)^{2}=0 ; & \text { that is } \\
x=y & \text { and } & x=0 & \text { and } & y=0 \quad \text { and } & z-y+5=0
\end{array}
$$

These can be satisfied simultaneously if $x=0, y=0$ and $z=-5$. This is therefore the only solution to the equation, which can be easily verified.