L.C.M. of $x^{3} - 1$ and $x^{4} + x^{2} + 1$ will be

(x - 1) (x^{2} + x + 1) (x^{2} - x + 1)

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(x - 1) (x^{2} + x + 1) (x^{2} - x - 1)

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(x - 1) (x^{2} + x - 1) (x^{2} - x + 1)

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(x + 1) (x^{2} + x + 1) (x^{2} - x + 1)

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Solution

The correct option is A $(x - 1) (x^{2} + x + 1) (x^{2} - x + 1)$
Given, $x^{3} - 1 = (x - 1) \times (x^{2} + x + 1)$
$x^{4} + x^{2} + 1 = (x^{2} + x + 1) \times (x^{2} - x + 1)$
Thus, the required L.C.M. is $(x - 1) \times (x^{2} + x + 1) \times (x^{2} - x + 1)$

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