1. Find all pairs $(r, s)$ of real numbers such that the zeros of the polynomials
$$
f(x)=x^{2}-2 r x+r
$$
and
$$
g(x)=27 x^{3}-27 r x^{2}+s x-r^{6}
$$
are all real and nonnegative.
1. Find all pairs $(r, s)$ of real numbers such that the zeros of the polynomials
$$
f(x)=x^{2}-2 r x+r
$$
and
$$
g(x)=27 x^{3}-27 r x^{2}+s x-r^{6}
$$
are all real and nonnegative.
Answers: $(0,0),(1,9)$
Solution:
Let $x_{1}, x_{2}$ be the zeros of $f(x)$, and let $y_{1}, y_{2}, y_{3}$ be the zeros of $g(x)$.

By Viete's relation,
$$
\begin{aligned}
x_{1}+x_{2} & =2 r \\
x_{1} x_{2} & =r
\end{aligned}
$$
and
$$
\begin{aligned}
y_{1}+y_{2}+y_{3} & =r \\
y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1} & =\frac{s}{27} \\
y_{1} y_{2} y_{3} & =\frac{r^{6}}{27}
\end{aligned}
$$

Note that
$$
\begin{aligned}
\left(\frac{x_{1}+x_{2}}{2}\right)^{2} \geq x_{1} x_{2} & \Rightarrow r^{2} \geq r \\
\frac{y_{1}+y_{2}+y_{3}}{3} & \geq \sqrt[3]{y_{1} y_{2} y_{3}} \\
\frac{r}{3} & \geq \sqrt[3]{\frac{r^{6}}{27}} \\
r & \geq r^{2}
\end{aligned}
$$

Hence $r=r^{2}$, and consequently $x_{1}=x_{2}$ and $y_{1}=y_{2}=y_{3}$. Moreover, $r=0,1$.
- If $r=0$, then $f(x)=x^{2}$ with $x_{1}=x_{2}=0$. And since $y_{1}=y_{2}=y_{3}$ with $y_{1}+y_{2}+y_{3}=0$, then ultimately $s=0$.
- If $r=1$, then $f(x)=x^{2}-2 x+1=(x-1)^{2}$ with $x_{1}=x_{2}=1$. And since $y_{1}=y_{2}=y_{3}$ with $y_{1}+y_{2}+y_{3}=1$ then $y_{1}=y_{2}=y_{3}=\frac{1}{3}$. Therefore $s=9$.

Thus, the possible ordered pairs $(r, s)$ are $(0,0)$ and $(1,9)$.