Given that, xy + yx = 1 x^2 + y^2 = xy x^2 + y^2 xy = 0 Now, x^3 + y^3 = (x+y)(x^2 + y^2 xy) #160; #160; #160; #160; #160; ...

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Algebraic Identities

If xy+yx=1;...

Question

If $\frac{x}{y} + \frac{y}{x} = 1; x^{3} + y^{3} = ?$

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Solution

Given that,
$\frac{x}{y} + \frac{y}{x} = 1$

$⟹ x^{2} + y^{2} = x y$

$⟹ x^{2} + y^{2} - x y = 0$

Now, $x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$

$= (x + y) (0)$

$= 0$ (Ans)

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Similar questions

If $\frac{x}{y} + \frac{y}{x} = - 1$ then find $x^{3} - y^{3}$

\frac{x}{y} + \frac{y}{x} = - 1, (x, y \neq 0)

x^{3} - y^{3}

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x^{3} + y^{3} + 3 x y = . . . . . . . . . . . .

x^{3} + y^{3} + 1 = 3 x y,

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x + y + 1.

\frac{x}{y} + \frac{y}{x} = 2, x \neq 0, y \neq 0

x^{3} - y^{3}

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