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Question

Find x3+y312xy+64, when x+y=4.

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Solution

(x+y)2=x2+y2+2xy
42=x2+y2+2xy
162xy=x2+y2
x2+y2=162xy
now,
x3+y3=(x+y)(x2+y2xy)
x3+y3=(4)(162xyxy)
x3+y3=(4)(163xy)
x3+y3=6412xy
x3+y364+12xy=0

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