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Implicit Differentiation

3 √3 x3 + y3 ...

Question

$3 \sqrt{3} x^{3} + y^{3}$ is equal to

A

(\sqrt{3} x + y) (3 x^{2} - \sqrt{3} x y + y^{2})

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B

(\sqrt{3} x - y) (\sqrt{3} x^{2} + \sqrt{3} x y + y^{2})

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C

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

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D

(\sqrt{3} x - y) (3 x^{2} - \sqrt{3} x y - y^{2})

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Solution

The correct option is A $(\sqrt{3} x + y) (3 x^{2} - \sqrt{3} x y + y^{2})$
Given, $3 \sqrt{3} x^{3} + y^{3} = (\sqrt{3} x)^{3} + y^{3}$
Since, $x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$
$∴ (\sqrt{3} x)^{3} + y^{3} = (\sqrt{3} x + y) ((\sqrt{3} x)^{2} + y^{2} - \sqrt{3} x y)$
$= (\sqrt{3} x + y) (3 x^{2} + y^{2} - \sqrt{3} x y)$
$= (\sqrt{3} x + y) (3 x^{2} - \sqrt{3} x y + y^{2})$
Option A is correct.

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