\(x^3-y^3-3xy\)
\(=x^3-y^3-3xy\left(x-y\right)\)
\(=\left(x-y\right)^3=1\)
Theo giả thiết:
\(x-y=1\Rightarrow x-1=y\Rightarrow\left(x-1\right)^3=y^3\Rightarrow x^3-3x^2+3x-1=y^3\Rightarrow x^3-y^3-3xy=3x^2-3x+1-3xy\)
\(\Rightarrow x^3-y^3-3xy=3x\left(x-1-y\right)+1=3x\left[\left(x-y\right)-1\right]+1=0+1=1\)
x-y=1=> (x-y)3=1
<=> x3- 3x3y - 3xy3 - y3
<=>x3 - y3 - 3xy(x+y)
=> x3-y3-3xy = 1