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