Question

if x/y+y/x=-1(where x ,y is not equal to zero),then the value of x3-y3 is​

Answer 1

Answer:

0

Step-by-step explanation:

Given :

To find the value of :

$x^3 - y^3$ if,

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

x ≠ 0 , y ≠ 0

Solution :

We know that,

$x^3 - y^3 = (x - y)(x^2 + y^2 + xy)$  ...(i)

The given equation meant that,

⇒ $\frac{x}{y} +\frac{y}{x} = -1$

⇒ $\frac{x \times x + y \times y}{xy} = -1$

⇒ $\frac{x^2 + y^2}{xy} = -1$

⇒ $x^2 + y^2 = -1 \times xy$

⇒ $x^2 + y^2 = -xy$

⇒ $x^2 + y^2 + xy = 0$  ...(ii)

By substituting the value from (ii) in (i),

We get,

⇒ $x^3 - y^3 = (x - y)(x^2 + y^2 + xy)$

⇒ $x^3 - y^3 = (x - y) \times 0$   (from (ii) )

⇒ $x^3 - y^3 = 0$