## Task A-1.2. (4 points)

Let $a$ be a real number. Determine the sum of all three solutions of the equation

$$
x^{3}-a^{2} x+a x-x+a^{2}-a=0
$$
## Solution.

The expression on the left side of the given equation can be factored as follows:

$$
\begin{aligned}
x^{3}-a^{2} x+a x-x+a^{2}-a & =\left(x^{3}-x\right)+a^{2}(-x+1)+a(x-1) & & 1 \text { point } \\
& =(x-1)\left(x^{2}+x-a^{2}+a\right) & & 1 \text { point } \\
& =(x-1)\left(\left(x^{2}-a^{2}\right)+(x+a)\right) & & \\
& =(x-1)(x+a)(x-a+1) & & 1 \text { point }
\end{aligned}
$$

It is now clear that the solutions to the equation $(x-1)(x+a)(x-a+1)=0$ are the numbers $x=1$, $x=-a$, and $x=a-1$, and their sum is 0.

Note: Since the coefficient of $x^{2}$ is 0, it follows from Vieta's formulas that the sum of the solutions of the given equation is zero.