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Question
If xy+yx=−1, where x,y≠0 then the value of (x3−y3) is
(a) 1
(b)−1
(c) 0
(d) 12
If xy+yx=−1, where x,y≠0 then the value of (x3−y3) is
(a) 1
(b)−1
(c) 0
(d) 12
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Solution
ANSWER:
(c) 0
xy+yx=−1
⇒x2+y2xy=−1
⇒ x2+y2=−xy
⇒ x2+y2+xy=0
(x3−y3)=(x+y) (x2+y2+xy)=0
Hence, (x3−y3)=0
ANSWER:
(c) 0
xy+yx=−1
⇒x2+y2xy=−1
⇒ x2+y2=−xy
⇒ x2+y2+xy=0
(x3−y3)=(x+y) (x2+y2+xy)=0
Hence, (x3−y3)=0
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