H.C.F. of $x^{3} - 1 a n d x^{4} + x^{2} + 1$ 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 C $x^{2} + x + 1$
Rewrite $x^{3} - 1$ as $x^{3} - 1^{3}$

Since $a^{3} - b^{3} = (a - b) (a^{2} + a + 1)$ , therefore,

$x^{3} - 1 = x^{3} - 1^{3} = (x - 1) (x^{2} + x + 1)$ ............(1)

Now consider $x^{4} + x^{2} + 1$ and factorise it as follows:

$x^{4} + x^{2} + 1 = (x^{4} + x^{2} + x^{2} + 1) - x^{2}$
$= (x^{4} + 2 x^{2} + 1) - x^{2}$
$= (x^{2} + 1)^{2} - (x)^{2}$ [as $(a + b)^{2} = a^{2} + b^{2} + 2 a b)$ ]
$= (x^{2} + 1 + x) (x^{2} + 1 - x)$ ............(2) [as $a^{2} - b^{2} = (a + b) (a - b)$ ]

Now, the highest common factor of equations 1 and 2 is $x^{2} + x + 1$ .

Hence, the H.C.F of $x^{3} - 1$ and $x^{4} + x^{2} + 1$ is $x^{2} + x + 1$ .

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Similar questions

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

x

\frac{X^{3} + X^{2} + X + 1}{X^{3} - X^{2} + X - 1}

=

\frac{X^{2} + X + 1}{X^{2} - X + 1}

X

ϵ

The simplest form of $\frac{(x^{3} - 1)}{(x^{2} - 1)}$ is

(x^{2} + x + 1)

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

L i m x \to 1 [{[\frac{4}{x^{2} - x^{- 1}} - \frac{{1 - 3 x + x}^{2}}{{1 - x}^{3}}]}^{- 1} + \frac{3 (x^{4} - 1)}{x^{3} - x^{- 1}}] =

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