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If -1 is a twice repeated root of the equation $a x^{3} + b x^{2} + c x + 1 = 0$ , then

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$b = 2 a + 1, c = a - 2$

$b = 2 a + 1, c = a + 1$

$b = 2 a + 1, c = a + 2$

$b = 2 a - 1, c = a + 2$

answer is C.

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Detailed Solution

-1 is the root of the equation $a x^{3} + b x^{2} + c x + 1 = 0$ then

$- a + b - c + 1 = 0$ and

-1 is a root of $3 a x^{2} + 2 b x + c = 0$ then

$3 a - 2 b + c = 0$

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