8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-3=0$. Find $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)$.
Answer: -1063.

Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+3$. Squaring, we get $x^{4}=$ $x^{2}+6 x+9=7 x+12$. Multiplying by $x$, we get $x^{5}=7 x^{2}+12 x=19 x+21$. Substituting the roots of the equation for $x$, we get: $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)=\left(19 x_{1}+21-20\right) \cdot\left(21 x_{2}+36-\right.$ $\left.2 x_{2}-35\right)=\left(19 x_{1}+1\right)\left(19 x_{2}+1\right)=361 x_{1} x_{2}+19\left(x_{1}+x_{2}\right)+1$. By Vieta's theorem $x_{1} x_{2}=$ $-3, x_{1}+x_{2}=1$. Substituting, we get $361 *(-3)+19+1=-1063$.

## Variant 1-b