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

$2$

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$3$

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$1$

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None

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Solution

The correct option is B
$3$

$a x^{2} + b x + c = 0$ and $b x^{2} + c x + a = 0$ has a common root, then

$\frac{x^{2}}{a b - c^{2}} = \frac{x}{c b - a^{2}} = \frac{1}{a c - b^{2}}$

$\frac{x^{2}}{a b - c^{2}} = \frac{1}{a c - b^{2}}$
$\Rightarrow x^{2} = \frac{a b - c^{2}}{a c - b^{2}} . . . . . . . . (1)$

$\frac{x}{c b - a^{2}} = \frac{1}{a c - b^{2}}$
$\Rightarrow x = \frac{c b - a^{2}}{a c - b^{2}}$

Using $(1)$
$\frac{a b - c^{2}}{a c - b^{2}} = {(\frac{c b - a^{2}}{a c - b^{2}})}^{2}$
$\Rightarrow (c b - a^{2})^{2} = (a b - c^{2}) (a c - b^{2})$
$\Rightarrow b^{2} c^{2} + a^{4} - 2 a^{2} b c = a^{2} b c - a b^{3} - a c^{3} + c^{2} b^{2}$

$\Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$

Suggest Corrections

Similar questions

a, b, c

a x^{2} + b x + c = 0,

b x^{2} + c x + a = 0

\frac{a^{3} + b^{3} + c^{3}}{a b c} =

a x^{2} + b x + c = 0

b x^{2} + c x + a = 0

a \neq 0

\frac{a^{3} + b^{3} + c^{3}}{a b c}

a x^{2} + b x + c = 0, b x^{2} + c x + a = 0

1

a, b, c

\frac{a^{3} + b^{3} + c^{3}}{3 a b c}

{a x}^{2} + b x + c = 0, {b x}^{2} + c x + a = 0

\frac{a^{3} + b^{3} + c^{3}}{a b c}

If $a x^{2} + b x + c = 0$ and $b x^{2} + c x + a = 0$ have a common root

a ≠ 0, then $\frac{a^{3} + b^{3} + c^{3}}{a b c} =$

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