11.2. Given different real numbers $a_{1}, a_{2}, a_{3}$ and $b$. It turned out that the equation $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)=b$ has three different real roots $c_{1}, c_{2}, c_{3}$. Find the roots of the equation $(x+$ $\left.+c_{1}\right)\left(x+c_{2}\right)\left(x+c_{3}\right)=b$.

(A. Antropov, K. Sukhov)
Answer: $-a_{1}, -a_{2}$, and $-a_{3}$.

First Solution. Since the polynomial $\left(x-a_{1}\right)\left(x-a_{2}\right)(x- \left.-a_{3}\right)-b$ has a leading coefficient of 1 and roots $c_{1}, c_{2}, c_{3}$, we have $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)-b=\left(x-c_{1}\right)\left(x-c_{2}\right)\left(x-c_{3}\right)$. Substituting $-x$ for $x$ in the last equation, we get $\left(-x-a_{1}\right)\left(-x-a_{2}\right)(- \left.-x-a_{3}\right)-b=\left(-x-c_{1}\right)\left(-x-c_{2}\right)\left(-x-c_{3}\right)$, which is equivalent to $\left(x+a_{1}\right)\left(x+a_{2}\right)\left(x+a_{3}\right)+b=\left(x+c_{1}\right)\left(x+c_{2}\right)\left(x+c_{3}\right)$. From the obtained equation, we see that the three roots of the equation $b=(x+ \left.+c_{1}\right)\left(x+c_{2}\right)\left(x+c_{3}\right)$ are the numbers $-a_{1}, -a_{2}, -a_{3}$.

Second Solution. By Vieta's formulas, the following relations hold:

$$
\begin{gathered}
c_{1}+c_{2}+c_{3}=a_{1}+a_{2}+a_{3} \\
c_{1} c_{2}+c_{2} c_{3}+c_{3} c_{1}=a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1} \\
c_{1} c_{2} c_{3}=a_{1} a_{2} a_{3}-b
\end{gathered}
$$

These equations can also be rewritten as:

$$
\begin{gathered}
\left(-a_{1}\right)+\left(-a_{2}\right)+\left(-a_{3}\right)=\left(-c_{1}\right)+\left(-c_{2}\right)+\left(-c_{3}\right) \\
a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}=c_{1} c_{2}+c_{2} c_{3}+c_{3} c_{1} \\
-a_{1} a_{2} a_{3}=-c_{1} c_{2} c_{3}-b
\end{gathered}
$$

from which it follows that the numbers $-a_{1}, -a_{2}$, and $-a_{3}$ are the roots of the equation $\left(x+c_{1}\right)\left(x+c_{2}\right)\left(x+c_{3}\right)-b=0$.

Remark. If we substitute the numbers $c_{1}, c_{2}$, and $c_{3}$ into the first equation, we get the three equations $\left(c_{1}-a_{1}\right)\left(c_{1}- \right. -a_{2}\left)\left(c_{1}-a_{3}\right)=b, \left(c_{2}-a_{1}\right)\left(c_{2}-a_{2}\right)\left(c_{2}-a_{3}\right)=b\right.$ and $\left(c_{3}-a_{1}\right)\left(c_{3}- \right. -a_{2}\left)\left(c_{3}-a_{3}\right)=b\right.$, from which it does not directly follow, for example, that $\left(-a_{1}+c_{1}\right)\left(-a_{1}+c_{2}\right)\left(-a_{1}+c_{3}\right)=b$.

Comment. A correct answer without justification: -1 point.

Algebraic transformations that do not lead to justification of the answer: no points added.