HCF of $x^{2} - y^{2}$ and $x^{3} - y^{3}$ is

x - y

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x^{3} - y^{3}

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(x^{2} - y^{2})

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(x + y) (x^{2} + x y + y^{2})

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Solution

The correct option is C $x - y$
Since, $x^{2} - y^{2} = (x + y) (x - y)$
$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$
$H . C . F . = (x - y)$
Option A is correct.

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Similar questions

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{(x - y)^{2} + 3 x y}] + [\frac{(x + y)^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

(i) x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2}) (i i) x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x 3 - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

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