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Question

Question 19
If xy+yx=1 (where x,y0 ),then the value x3y3 is

A) 1
B) -1
C) 0
D) 12


Solution

Given xy+yx=1
 x2+y2xy=1
 x2+y2=xy
 x2+y2+xy=0
Now, x3y3=(xy)(x2+xy+y2) ---(i)
[using identity, a3b3=(ab)(a2+ab+b2)]
=(xy)×0=0                  [from Eq (i)]
Hence, the answer is C.

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