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Question

If x+y+4=0, find the value of x2+y212xy+64

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Solution

It is given that x+y+4=0 or x+y=4

We know the identity (a+b)3=a3+b3+3ab(a+b)

Now consider x+y=4 and take cube on both sides and then appy the above identity as follows:

(x+y)3=(4)3x3+y3+3xy(x+y)=64x3+y3+3xy(4)=64x3+y312xy+64=0

Hence, x3+y312xy+64=0


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