If $a, b, c$ are non-zero real numbers and $a x^{2} + b x + c = 0,$ $b x^{2} + c x + a = 0$ have one root in common, then $\frac{a^{3} + b^{3} + c^{3}}{a b c} =$

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Solution

The correct option is C $3$
$a x^{2} + b x + c = 0, b x^{2} + c x + a = 0$ have one root in common
$\Rightarrow (b c - a^{2})^{2} = (a b - c^{2}) (a c - b^{2})$
$\Rightarrow b^{2} c^{2} + a^{4} - 2 a^{2} b c = a^{2} b c - a b^{3} - a c^{3} + b^{2} c^{2}$
$\Rightarrow a^{4} + a b^{3} + a c^{3} = 3 a^{2} b c$
$\Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$
$∴ \frac{a^{3} + b^{3} + c^{3}}{a b c} = 3$

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