Question

ind the value of x^3 - y^3 , if x/y + y/x = (-1) and (x,y is not equall to 0) please answer no spamming

Answer 1

Answer:

0

Step-by-step explanation:

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

Taking LCM, we get

$\frac{ {x}^{2} \: + \: {y}^{2} }{xy} = \: - 1$

${x}^{2} \: + \: {y}^{2} = - xy$

${x}^{2} \: + \: {y}^{2} \: + \: xy = 0$

Multiply (x - y) on both sides.

$(x - y)( {x}^{2} + xy + {y}^{2} ) \: = \: 0$

${x}^{3} \: - \: {y}^{3} \: = \: 0$

Answer 2

Step-by-step explanation:

Given :-

We know that (a^3-b^3=(a+b) (a^2+b^2-2ab).

So, x^3-y^3=(x+y) (x^2+y^2-2xy).

it is given that

x/y+y/x=-1

x^2+y^2=-xy

x^+y^2+xy=0

x^2+xy+y^2=0

Now,Multiplying by (x-y) by both's side.

(x-y) (x^2+xy+y^2)=x-y×0

By eq. 1 we get,

(x^3-y^3) = 0.

Proved.