The GCD of $(x^{3} - 1)$ and $x^{4} - 1$ is

x^{3} - 1

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x^{3} + 1

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x + 1

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x - 1

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Solution

The correct option is D $x + 1$
GCD = Greatest Common Divisor.

$x^{3} - 1 = (x - 1) (x^{2} + x y + y 2)$ {identity $a^{3} - b^{3}$ and $a^{2} - b^{2}$ }

$x^{4} - 1 = (x^{2} - 1) (x^{2} + 1)$

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

GCD $= (x - 1)$

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Similar questions

The GCD of x³-x²+x-1 and x⁴-1 is _____________.

x^{3} - x^{2} + x - 1, x^{4} - 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^{4} + \frac{1}{x^{4}} = 119

x > 1

x^{3} - \frac{1}{x^{3}}

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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