If two quadratic equations
$x^{2} + b x - 1 = 0 & x^{2} + x + b = 0$ have a root in common, then $b^{3} + 3 b =$

1

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0

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Solution

The correct option is C $0$
Let $α$ be the common root of given equations, then
$α^{2} + b α - 1 = 0 \dots (1)$
and $α^{2} + α + b = 0 \dots (2)$
Subtracting $(2) from (1),$ we get
$(b - 1) α - (b + 1) = 0$
$\Rightarrow α = \frac{b + 1}{b - 1}$
Substituting this value of $α$ in equation $(1),$ we get
$(\frac{b + 1}{b - 1})^{2} + b (\frac{b + 1}{b - 1}) - 1 = 0 \Rightarrow (b + 1)^{2} + b (b + 1) (b - 1) - (b - 1)^{2} = 0 \Rightarrow b^{2} + 2 b + 1 + b (b^{2} - 1) - b^{2} - 1 + 2 b = 0 \Rightarrow b^{3} + 3 b = 0$

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