The H.C.F. of $x^{3} + y^{3}$ and $x^{2} - x y + y^{2}$ is

x + y

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x^{2} - x y + y^{2}

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x^{3} + y^{3}

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{(x + y)}^{3}

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Solution

The correct option is B $x^{2} - x y + y^{2}$
Consider, $(x + y)^{3} = (x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2})$

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

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

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

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

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

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

$∴$ H.C.F. of $(x^{3} + y^{3})$ and $(x^{2} - x y + y^{2})$ is $(x^{2} - x y + y^{2})$ .

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Similar questions

[\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}}

[\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}}

(x + y) + (x^{2} + x y + y^{2}) + (x^{3} + x^{2} y + x y^{2} + y^{3}) + . . . n

=

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

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

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