HCF of $6 x^{2} y (x - y) (x^{2} + x y + y^{2}), 18 x y^{2} (x - y)^{2}$ and $12 x (x - y)^{3}$ is

6 x (x - y)

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6 (x - y)

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6 x

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x (x - y)

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Solution

The correct option is A $6 x (x - y)$
We have,
$\Rightarrow 6 x^{2} y (x - y) (x^{2} + x y + y^{2}), 18 x y^{2} (x - y)^{2}, 12 x (x - y)^{3}$
$\Rightarrow 6 x (x - y) \times x y (x^{2} + x y + y^{2}), 6 x (x - y) \times 3 y^{2} (x - y), 6 x (x - y) \times 2 (x - y) (x - y)$

Hence, the $H C F$ is $6 x (x - y)$ .

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Similar questions

6 x^{2} (x - y) (x^{2} + x y + y^{2}), (x - y)^{2}

12 x (x - y)^{3}

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

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

{(x + y + z)}^{2} - x^{5} - y^{5} - z^{5} = 5 (y + z) (z + x) (x + y) (x^{2} + y^{2} + z^{2} + y z + z x + x y)

(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})

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