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

x - 1

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x^{2} + x + 1

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x^{2} - x + 1

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x^{2} - x - 1

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Solution

The correct option is B $x^{2} + x + 1$
$x^{3} - 1 = x^{3} - 1^{3}$
$= (x - 1) (x^{2} + x + 1)$
$x^{4} + x^{2} + 1 = (x^{4} + 2 x^{2} + 1) - x^{2}$
$= (x^{2} + 1)^{2} - (x)^{2}$
$= (x^{2} + 1 + x) (x^{2} + 1 - x)$
$= (x^{2} + x + 1) (x^{2} - x + 1)$
Hence, H.C.F. of $x^{3} - 1$ and $x^{4} + x^{2} + 1$ is $x^{2} + x + 1$ .

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Similar questions

x^{3} - 1

x^{4} + x^{2} + 1

If $x + \frac{1}{x} = 3$ , calculate $x^{2} + \frac{1}{x^{2}}, x^{3} + \frac{1}{x^{3}} a n d x^{4} + \frac{1}{x^{4}}$ .

(x^{2} + x + 1)

(x^{4} + x^{2} + 1)

x \neq 0

x + \frac{1}{x} = 2

x^{2} + \frac{1}{x^{2}} = x^{3} + \frac{1}{x^{3}} = x^{4} + \frac{1}{x^{4}}

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