Problem 9.1. Let $f(x)=x^{2}+(2 a-1) x-a-3$, where $a$ is a real parameter.

a) Prove that the equation $f(x)=0$ has two distinct real roots $x_{1}$ and $x_{2}$.

b) Find all values of $a$ such that $x_{1}^{3}+x_{2}^{3}=-72$.

Peter Boyvalenkov
9.1. a) The discriminant of $f(x)$ is equal to $4 a^{2}+13>0$.

b) We consecutively have

$$
\begin{aligned}
-72 & =x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right] \\
& =(1-2 a)\left(4 a^{2}-a+10\right)=-8 a^{3}+6 a^{2}-21 a+10
\end{aligned}
$$

Therefore the required values of $a$ are the real solutions of the equation

$$
8 a^{3}-6 a^{2}+21 a-82=0
$$

Since $a=2$ is a solution of this equation, we obtain

$$
8 a^{3}-6 a^{2}+21 a-82=(a-2)\left(8 a^{2}+10 a+41\right)
$$

The equation $8 a^{2}+10 a+41=0$ has no real roots. So the only solution of the problem is $a=2$.