6. The polynomial $p(x)=x^{2}-3 x+1$ has zeros $r$ and $s$ and a quadratic polynomial $q(x)$ has leading coefficient 1 and zeros $r^{3}$ and $s^{3}$. Find $q(1)$.
Answer: -16
Solution: We have $r+s=3$ and $r s=1$. By Vieta's formulas, we compute
$$
r^{3}+s^{3}=(r+s)^{3}-3 r s(r+s)=27-3(1)(3)=18
$$
and $r^{3} s^{3}=1$. Thus, we obtain $q(x)=\left(x-r^{3}\right)\left(x-s^{3}\right)=x^{2}-\left(r^{3}+s^{3}\right) x+r^{3} s^{3}=x^{2}-18 x+1$. Hence, we get $q(1)=-16$.