Question
If $a, b, c$ are all non-zero and $a+b+c=0$, prove that $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}=3$.
✓
Answer
We have $a, b, c$ are all non-zero and $a+b+c=0$, therefore
$a^3+b^3+c^3=3 a b c$
Now, $\frac{\mathrm{a}^2}{\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{ab}}=\frac{\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3}{\mathrm{abc}}=\frac{3 \mathrm{abc}}{\mathrm{abc}}=3$
$a^3+b^3+c^3=3 a b c$
Now, $\frac{\mathrm{a}^2}{\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{ab}}=\frac{\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3}{\mathrm{abc}}=\frac{3 \mathrm{abc}}{\mathrm{abc}}=3$
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