If $a x^{2} + b x + c = 0 a n d b x^{2} + c x + a = 0$ have a common root and a, b, c are non-zero real numbers, then find the value of $\frac{a^{3} + b^{3} + c^{3}}{a b c}$

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Solution

The correct option is C
3

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

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

Let $α$ be the common root,

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

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

$\Rightarrow$ $a^{2} b c - a c^{3} - a b^{3} + b^{2} c^{2} = b^{2} c^{2} - 2 a^{2} b c + a^{4}$

$\Rightarrow$ $a c^{3} + a^{4} + a b^{3} = 3 a^{2} b c$

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

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Similar questions

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, 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

\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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