find the value of x^3-y^3 if x/y+y/x=-1
xy are not 0
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Step-by-step explanation:
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.
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