Question

if 1/x+y=1/x+1/y (x≠0,y≠0,z≠y) then, find the value of x³-y³?

Answer 1

Hi,

Here's the solution :-

1/(x+y) = 1/x + 1/y

=> 1/ (x + y) = (x + y)/xy

=>( x+y)^2 = xy

=> x^2 + y^2 + 2xy = xy

=> x^2 + xy + y^2 = 0

According to question,

x^3 - y^3

= (x - y)( x^2 + xy + y^2)

= (x - y) x 0

= 0


Hope that you are clear ^_^

Answer 2

Step-by-step explanation:

hope this helps you

and you get your answer