if x+y+1=0, prove that x^3+y^3+1=3xy
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x+y=(-1)
x^3+y^3+1=(x+y)(x^2-xy+y^2)+1=(-1)*[(x+y)^2-3xy]+1
= -1*(1-3xy)+1
=3xy
x^3+y^3+1=(x+y)(x^2-xy+y^2)+1=(-1)*[(x+y)^2-3xy]+1
= -1*(1-3xy)+1
=3xy
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