If $α, β, γ$ are the roots of the equation $x^{3} + x + 1 = 0$ , then the value of $\prod (\frac{1 + α}{1 - α})$ , is:

- 7

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{3}$
$α, β, γ$ are roots of $x^{3} + x + 1 = 0$
Then $α + 1, β + 1, γ + 1$ are roots of ${(x - 1)}^{3} + (x - 1) + 1 = 0$
$\Rightarrow x^{3} - 1 - 3 x^{2} + 3 x + x - 1 + 1 = 0 \Rightarrow x^{3} - 3 x^{2} + 4 x - 1 = 0$
And $(α + 1) (β + 1) (γ + 1) = 1$
Similarly $α - 1, β - 1, γ - 1$ are roots of ${(x + 1)}^{3} + (x + 1) + 1 = 0$
$\Rightarrow x^{3} + 1 + 3 x^{2} + 3 x + x + 1 + 1 = 0 \Rightarrow x^{3} + 3 x^{2} + 4 x + 3 = 0$
$∴ (α - 1) (β - 1) (γ - 1) = - 3$
Now $\prod (\frac{1 + α}{1 - α}) = - (\frac{1}{- 3}) = \frac{1}{3}$

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