If $x + y - 1 = 0$ , then $x^{3} + y^{3} - 1$ is equal to

x^{2} + y^{2} - 1

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

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x^{2} + x y + y^{2}

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- 3 x y

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Solution

The correct option is D $- 3 x y$
We are given that $x + y - 1 = 0$ which is same as $x + y = 1$ .
Taking cubes on both sides of $x + y = 1$ , we get:
$(x + y)^{3} = 1^{3} \Rightarrow x^{3} + y^{3} + 3 x y (x + y) = 1 (∵ (a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b)) \Rightarrow x^{3} + y^{3} + (3 x y \times 1) = 1 (G i v e n : x + y = 1) \Rightarrow x^{3} + y^{3} + 3 x y = 1 \Rightarrow x^{3} + y^{3} - 1 = - 3 x y$

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[\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}}

∣ x ∣< 1

∣ y ∣< 1

(x + y) + (x^{2} + x y + y^{2}) + (x^{3} + x^{2} y + x y^{2} + y^{3}) + . . . . . \infty

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

y - x = 1

\frac{x y}{x^{2} + y^{2}}

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