If $x = 1 + i$ is a root of the equation $x^{3} - i x + 1 - i = 0$ , then the other real root is

0

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1

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- 1

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None of the above.

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Solution

The correct option is C $- 1$
Given equation is
$x^{3} - i x + 1 - i = 0$
$x^{3} + 1 - i (x + 1) = 0$
$(x + 1) (x^{2} - x + 1 - i) = 0$
$∴$ Real root of given equation is $x = - 1$

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