Bài 5:
a: Ta có: \(A=x^3+y^3+3xy\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=\left(x+y\right)^2\)
=1
b: Ta có: \(B=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\cdot\left(x^2-xy+y^2\right)-3\cdot\left(-2xy\right)\)
\(=2x^2-2xy+2y^2+6xy\)
\(=2\cdot\left(x^2+2xy+y^2\right)\)
=2