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Polynomial Expressions

Question 19If...

Question

Question 19
If $\frac{x}{y} + \frac{y}{x} = - 1$ (where $x, y \neq 0$ ),then the value $x^{3} - y^{3}$ is

A) 1
B) -1
C) 0
D) $\frac{1}{2}$

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Solution

Given $\frac{x}{y} + \frac{y}{x} = - 1$
$\Rightarrow \frac{x^{2} + y^{2}}{x y} = - 1$
$\Rightarrow x^{2} + y^{2} = - x y$
$\Rightarrow x^{2} + y^{2} + x y = 0$
Now, $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ ---(i)
[using identity, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ ]
$= (x - y) \times 0 = 0$ [from Eq (i)]
Hence, the answer is C.

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Similar questions

Q. Question 19
If

\frac{x}{y} + \frac{y}{x} = - 1

x, y \neq 0

),then the value

x^{3} - y^{3}

is

A) 1
B) -1
C) 0
D)

\frac{1}{2}

\frac{x}{y} + \frac{y}{x} = - 1

, where x ≠ 0 and y ≠ 0, then the value of (x³ − y³) is
(a) 1
(b) −1
(c) 0
(d)

\frac{1}{2}

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

, the value of x³– y³ is
(a) 1
(b) –1
(c) 0
(d)

\frac{1}{2}

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

, the value of x³+ y³ is
(a) 1
(b) –1
(c) 0
(d)

- \frac{1}{2}

If $\frac{x}{y} + \frac{y}{x} = - 1$ , where $x, y \neq 0$ then the value of $(x^{3} - y^{3})$ is
(a) $1$

(b) $- 1$

(c) $0$

(d) $\frac{1}{2}$

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