12. Suppose the real numbers $x$ and $y$ satisfy the equations
$$
x^{3}-3 x^{2}+5 x=1 \text { and } y^{3}-3 y^{2}+5 y=5
$$

Find $x+y$.
12. Answer: 2
Solution. From $x^{3}-3 x^{2}+5 x=1$, we have
$$
(x-1)^{3}+2(x-1)=-2,
$$
and from $y^{3}-3 y^{2}+5 y=5$, we have
$$
(y-1)^{3}+2(y-1)=2 .
$$

Thus
$$
\begin{array}{c}
0=(x-1)^{3}+2(x-1)+(y-1)^{3}+2(y-1) \\
=(x+y-2)\left((x-1)^{2}-(x-1)(y-1)+(y-1)^{2}\right)+2(x+y-2) \\
=(x+y-2)\left(2+(x-1)^{2}-(x-1)(y-1)+(y-1)^{2}\right) .
\end{array}
$$

For any real numbers $x$ and $y$, we always have
$$
(x-1)^{2}-(x-1)(y-1)+(y-1)^{2} \geq 0
$$
and thus $x+y-2=0$, implying that $x+y=2$.