## Task B-1.5.

For real numbers $a, b, c$, which are not equal to zero, it holds that $a+b+c=0$. What is

$$
\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b} ?
$$
## Solution.

If $a+b+c=0$, then $c=-(a+b)$.

The given expression is reduced to a common denominator and we substitute $c=-(a+b)$. We get

$$
\begin{array}{rlrl}
\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b} & =\frac{a^{3}+b^{3}-(a+b)^{3}}{-a b(a+b)} & & 1 \text { point } \\
& =\frac{(a+b)\left(a^{2}-a b+b^{2}\right)-(a+b)^{3}}{-a b(a+b)} & & 1 \text { point } \\
& =\frac{a^{2}-a b+b^{2}-a^{2}-2 a b-b^{2}}{-a b} & & 2 \text { points } \\
& =\frac{-3 a b}{-a b}=3 . & 1 \text { point }
\end{array}
$$