# Problem 3. (3 points)

It is known that $a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c=30$ and $a^{2}+b^{2}+c^{2}=13$.

Find $a+b+c$.
Answer: 5

## Solution:

$(a+b+c)^{3}-(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)=2\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c\right)$. Let $a+b+c$ be $x$, substitute the known values of the expressions from the condition, and we get the equation $x^{3}-13 x-60=0$.

It is not hard to notice that the number 5 is a root of this equation. Dividing $x^{3}-13 x-60$ by $x-5$, we get $x^{2}+5 x+12$. This quadratic has no roots, so 5 is the only possible answer.