Problem 8.8.1. Different positive numbers $a, b, c$ are such that

$$
\left\{\begin{array}{l}
a^{2}+b c=115 \\
b^{2}+a c=127 \\
c^{2}+a b=115
\end{array}\right.
$$

Find $a+b+c$.
Answer: 22.

Solution. Subtract the third equation from the first and transform:

$$
\begin{gathered}
\left(a^{2}+b c\right)-\left(c^{2}+a b\right)=0 \\
a^{2}-c^{2}+b c-a b=0 \\
(a-c)(a+c)+b(c-a)=0 \\
(a-c)(a+c-b)=0
\end{gathered}
$$

By the condition $a \neq c$, therefore $b=a+c$. Now add the first two equations:

$$
\begin{gathered}
\left(a^{2}+b c\right)+\left(b^{2}+a c\right)=115+127 \\
\left(a^{2}+b c\right)+\left(b^{2}+a c\right)=242
\end{gathered}
$$

Substitute $b=a+c$ into the resulting equation:

$$
\begin{aligned}
242 & =a^{2}+b c+b^{2}+a c=a^{2}+(a+c) c+(a+c)^{2}+a c= \\
& =a^{2}+a c+c^{2}+(a+c)^{2}+a c=2(a+c)^{2}
\end{aligned}
$$

from which $(a+c)^{2}=121$. Since the numbers $a$ and $c$ are positive and $b=a+c$, we get that $a+c=11$ and $a+b+c=22$.