Question 19 If x-y-y-x-1 where x- y-0 -then the value x3-y3 isA 1B -1C 0D 1-2

Given x/y+y/x= 1 ⇒ x^2+y^2/xy= 1 ⇒ nbsp;x^2+y^2= xy ⇒ nbsp;x^2+y^2+xy=0 Now, x^3 y^3=(x y)(x^2+xy+y^2) —(i) [using identity, a^3 b^3=(a b)(a^2+ab+b^2)] =( ...

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Standard X

Mathematics

Factor of a Polynomial

Question 19 I...

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

(where

x, y \neq 0

),then the value

x^{3} - y^{3}

is

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

\frac{1}{2}

Q. If

\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}

Q. If

\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}

Q. If

\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) $- 1$

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

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Factor of a Polynomial

Standard X Mathematics

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Question 19 If xy+yx=−1 (where x,y≠0 ),then the value x3−y3 is A) 1 B) -1 C) 0 D) 12

Given xy+yx=−1 ⇒ x2+y2xy=−1 ⇒ x2+y2=−xy ⇒ x2+y2+xy=0 Now, x3−y3=(x−y)(x2+xy+y2) ---(i) [using identity, a3−b3=(a−b)(a2+ab+b2)] =(x−y)×0=0 [from Eq (i)] Hence, the answer is C.

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) $\frac{1}{2}$