The HCF of $x^{4} - y^{4} a n d x^{6} - y^{6}$ is

x^{2} - y^{2}

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x^{2} + y^{2}

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x^{3} + y^{3}

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x^{3} - y^{3}

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Solution

The correct option is A $x^{2} - y^{2}$
$x^{4} - y^{4} = (x^{2} - y^{2}) (x^{2} + y^{2})$
$= (x - y) (x + y) (x^{2} + y^{2})$

$x^{6} - y^{6} = {(x^{3})}^{2} - {(y^{3})}^{2} = (x^{3} - y^{3}) (x^{3} + y^{3})$
$= (x - y) (x^{2} + x y + y^{2}) (x + y) (x^{2} - x y + y^{2})$

$∴$ Reqd HCF = $(x - y) (x + y) = x^{2} - y^{2}$

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Similar questions

x^{2} - y^{2}

x^{3} - y^{3}

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

\frac{x^{2} + y^{2}}{x^{2} - y^{2}} = \frac{17}{8}

\frac{x^{3} + y^{3}}{x^{3} - y^{3}}

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0

[\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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