8. (10 points) Real numbers $a, b, x$ and $y$ are such that

$$
\left\{\begin{array}{l}
a x+b y=r-t \\
a x^{2}+b y^{2}=r \\
a x^{3}+b y^{3}=r+t \\
a x^{4}+b y^{4}=r+t^{2}
\end{array}\right.
$$

Find $a x^{2}+b y^{2}$.
Solution: Notice that

$$
\begin{gathered}
a x^{3}+b y^{3}=\left(a x^{2}+b y^{2}\right) \cdot(x+y)-x y \cdot(a x+b y) \\
a x^{4}+b y^{4}=\left(a x^{3}+b y^{3}\right) \cdot(x+y)-x y \cdot\left(a x^{2}+b y^{2}\right)
\end{gathered}
$$

We obtain a system of two equations:

$$
\left\{\begin{array}{l}
r \cdot(x+y)-(r-t) \cdot x y=r+t \\
(r+t) \cdot(x+y)-r \cdot x y=r+t^{2}
\end{array}\right.
$$

Solving it for $x y$ and $x+y$, we find

$$
\left\{\begin{array}{l}
x+y=\frac{2 r}{t}+t-r \\
x y=\frac{2 r}{t}-r+1
\end{array}\right.
$$

We can write similarly

$$
\begin{gathered}
a x^{5}+b y^{5}=\left(a x^{4}+b y^{4}\right) \cdot(x+y)-x y \cdot\left(a x^{3}+b y^{3}\right)= \\
=\left(r+t^{2}\right) \cdot\left(\frac{2 r}{t}+t-r\right)-\left(\frac{2 r}{t}-r+1\right) \cdot(r+t)= \\
=\frac{2 r^{2}}{t}+r t-r^{2}+2 r t+t^{3}-t^{2} r-\frac{2 r^{2}}{t}+r^{2}-r-2 r+t r-t= \\
=4 r t+t^{3}-t^{2} r-3 r-t=3 r \cdot(t-1)-r t \cdot(t-1)+t \cdot(t+1) \cdot(t-1)= \\
=(t-1) \cdot\left(t^{2}+t \cdot(1-r)+3 r\right)
\end{gathered}
$$

We obtained the final answer without calculating the actual numbers $x, y, a$, and $b$.

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 24 | 24 | 40 | 132 | 33 | 144 | 24 | 3 | 6 | 280 |