Tính
\(A=x^3-3xy-y^3\)
\(A=x^3-y^3-3xy\)
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(A=x^2+xy+y^2-3xy\)
\(A=x^2-2xy+y^2\)
\(A=\left(x-y\right)^2\)
\(A=1^2\)
\(A=1\)
Vậy A=1
\(A=x^3-3xy-y^3=x^3-3xy\left(x-y\right)-y^3\) ( vi x - y = 1)
\(=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=1^3=1\)