CBSE BOARD XII, asked by sahisingh4254, 9 months ago

Xy(x+y)=1 1/x³y³-x³-y³=?

Answers

Answered by anantrajusharma
0

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

JAI SHREE KRISHNA

Answered by shj0570515
0

Answer:

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 eq 1 we get,

x³-y³=0

Hence x³-y³=0

Explanation:

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