Factorise:
$x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} - 125 = 0$

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Solution

$x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} - 125 = 0$

$\Rightarrow (x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}) = 125$

$\Rightarrow [x^{3} + 3 x y (x + y) + y^{3}] = 125$

Using the identity, $a^{3} + 3 a b (a + b) + b^{3} = {(a + b)}^{3}$
we have

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

$\Rightarrow {(x + y)}^{3} = 5^{3}$

$\Rightarrow {(x + y)}^{3} - 5^{3}$ is of the form $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$

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

$= (x + y - 5) (x^{2} + y^{2} + 2 x y + 5 x + 5 y + 25)$

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Similar questions

x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} - 8

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The expansion of ${(x + y)}^{3}$ is $x^{3} + y^{3} + 3 x^{2} y - 3 x y^{2}$

x^{3} - 3 x y^{2} + 2 = 0

3 x^{2} y - y^{3} - 2 = 0

x^{3} - y^{3} + 3 x y^{2} - 3 x^{2} y + 1 = 0

(0, 1)

\frac{d y}{d x}

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