7. the polynomial $P(x)=x^{3}-2 x^{2}-x+1$ has the three real zeros $a>b>c$. Find the value of the expression

$$
a^{2} b+b^{2} c+c^{2} a
$$

## Solution
We set $A=a^{2} b+b^{2} c+c^{2} a$ and $B=a^{2} c+c^{2} b+b^{2} a$. Because $a>b>c$ applies

$$
A-B=(a-c)(c-b)(b-a)>0
$$

so $A$ is greater than $B$. In the following, we use the terms

$$
\begin{aligned}
u & =a+b+c \\
v & =a b+b c+c a \\
w & =a b c
\end{aligned}
$$

according to Vieta's theorem, $u=2, v=-1, w=-1$. Now applies

$$
\begin{aligned}
A+B & =a(a b+a c)+b(b c+b a)+c(c b+c a) \\
& =(a+b+c)(a b+b c+c a)-3 a b c=u v-3 w=1
\end{aligned}
$$

Using the identity $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ further follows

$$
\begin{aligned}
A \cdot B & =\left(a^{3}+b^{3}+c^{3}-3 a b c\right) a b c+\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3 a^{2} b^{2} c^{2}\right)+9 a^{2} b^{2} c^{2} \\
& =u\left(u^{2}-3 v\right) w+v\left(v^{2}-3 u w\right)+9 w^{2}=-12
\end{aligned}
$$

Again according to Vieta, $A$ and $B$ are therefore the two zeros of the polynomial $Q(x)=x^{2}-x-12=(x+3)(x-4)$. Because $A>B$ this results in the value $A=4$.