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ddebasish mallick

· started a discussion

· 1 Months ago

context of question is wrong (Question: 143206.0 )

the condition is given X not equals to y.
so if we get cube of x - cube of y = 0
that means x=y ultimately the condition is getting wrong.
hence the question should remove the clause of x not equals to y

Question:
If \(\cfrac{1}{x+y}\) = \(\cfrac{1}{x}\) + \(\cfrac{1}{y}\) (x≠0, y≠0)then, the value of x3 - y3 is: 
Options:
A) 0
B) 1
C) -1 
D) 2
Solution:

\(\cfrac{1}{x+y}\) = \(\cfrac{1}{x}\) + \(\cfrac{1}{y}\) = \(\cfrac{y+x}{xy}\)

⇒ (x+y)2 = xy ⇒ x2 + 2xy + y2 = xy

⇒ x2 + xy + y2 = 0

∴ x3 – y3 = (x – y)(x2 +xy + y2) = 0

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