Question

If (x/y+y/x)=-1 where x,y is not equal to zero ,then the value of x³-y³is *

a) 1

b) -1

c) 0

d) 1/2

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Answer 1

Answer:

We know that,

x³-y³=(x-y)(x²+xy+y²)........................1

Given that,

x/y+y/x=-1

(x²+y²)/xy=-1

x²+y²=-xy

x²+y²+xy=0

(x²+xy+y²)=0

Multyplying by (x-y) on both sides we get,

(x-y)(x²+xy+y²)=(x-y)×0

By eq1 we get,

x³-y³=0

Hence x³-y³=0.

or

x/y + y/x = –1

=> x²+y²+xy = 0

x³–y³ = (x–y) (x²+y²+xy) = (x–y) × 0 = 0