$(x + y)^{3} - (x - y)^{3}$ can be factorised as

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

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

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

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

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Solution

The correct option is B $2 y (3 x^{2} + y^{2})$
$(x + y)^{3} = x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2}$
$(x - y)^{3} = x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}$
Subtracting,
$(x + y)^{3} - (x - y)^{3} = 2 y^{3} + 6 x^{2} y$
$= 2 y (3 x^{2} + y^{2})$
So, $B$ is correct

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Similar questions

$(x + y)^{3} - (x - y)^{3}$ can be factorized as

[\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}}

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