Factorisation of $81 x^{3} - x y^{2}$ gives

$x (9 x - y) (9 x + y)$

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$x^{2} (9 x + y)$

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$x (9 y - x) (9 x + y)$

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$x (9 x - y) (9 x - 1)$

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Solution

The correct option is A
$x (9 x - y) (9 x + y)$

We need to factorise the given expression.
$81 x^{3} - x y^{2}$
$= x (81 x^{2} - y^{2})$ (taking $x$ common)
$= x [(9 x)^{2} - (y)^{2}]$
(converting in the form $a^{2} - b^{2}$ )
$= x [(9 x - y) (9 x + y)] = x (9 x - y) (9 x + y)$

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Similar questions

81 x^{3} - x y^{2}

81 x^{3} - 3 y^{3}

Using the factorisation of $x y^{2} - x z^{2}$ , evaluate the value of $6 \times 3^{2} - 6 \times 2^{2}$ .

x y^{2} - x z^{2}

40 \times {5.5}^{2} - 40 \times {4.5}^{2}

$Factorise x y^{2} - x z^{2}, Hence, find the value of : (i) 9 \times 8^{2} - 9 \times 2^{2} (i i) 40 \times {5.5}^{2} - 40 \times {4.5}^{2}$

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