If $x + y - 1 = 0$ , prove that $x^{3} + y^{3} + 3 x y = 1$

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Solution

$x + y - 1 = 0$
$\Rightarrow x + y = 1 \dots (i)$

Now, on taking cube both sides of the equation $(i)$ we get,
$\begin{matrix} (x + y)^{3} & = 1 x^{3} + y^{3} + 3 x y (x + y) & = 1 x^{3} + y^{3} + 3 x y & = 1 \end{matrix}$

Hence proved.

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