C1: \(B=x^3+3xy+y^3\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y-1\right)\)
Thay \(x+y=1\)ta được:
\(B=1^3-3xy\left(1-1\right)=1\)
C2: \(x+y=1\)\(\Rightarrow\)\(x=1-y\)
\(B=x^3+3xy+y^3=\left(1-y\right)^3+3\left(1-y\right)y+y^3\)
\(=1-3y+3y^2-y^3+3y-3y^2+y^3=1\)