ta có : \(x^3-y^3-3xy=\left(x-y\right)^3+3xy\left(x-y\right)-3xy\)
\(=\left(x-y\right)^3+3xy\left(x-y-1\right)=1^3+3xy\left(1-1\right)\)
\(=1+3xy\left(0\right)=1+0=1\)
vậy \(x^3-y^3-3xy=1\) với \(x-y=1\)
\(x^3-y^3-3xy=x^3-y^3-3xy.1\)
mà x- y =1 nên
\(x^3-y^3-3xy=x^3-y^3-3xy\left(x-y\right)\)
\(=x^3-y^3-3x^2y+3xy^2\)
=\(\left(x-y\right)^3\)
\(=1^3\)
= 1
Với x - y= 1 => \(x^3-y^3-3xy=1\)
\(x^3-y^3-3xy\\ =\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\\ =\left(x^2+xy+y^2\right)-3xy\\ =x^2+xy+y^2-3xy\\ =x^2+\left(xy-3xy\right)+y^2\\ =x^2-2xy+y^2\\ =\left(x-y\right)^2\\ =1^2\\ =1\)
Vì x-y = 1=>x = y+1
=> \(x^3-y^3-3xy=\left(y+1\right)^3-y^3-3\left(y+1\right)y\\ y^3+3y^2+3y+1-y^3-3y^2-3y=1\)
vậy x^3-y^3-3xy=1 nếu x-y = 1