Question

Giải phương trình nghiệm nguyên \(x^3+y^3+1=3xy\)

Answer 1

\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+1=3xy\)

\(\Leftrightarrow\left(x+y\right)^3+1=3xy\left(x+y+1\right)\)

\(\Leftrightarrow\left(x+y+1\right)\text{[}\left(x+y\right)^2-\left(x+y\right)+1\text{]}=3xy\left(x+y+1\right)\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2-xy+y^2-x-y+1\right)=0\)

Xét \(x,y\ge0\)

\(\Rightarrow x^3+y^3+1\ge3\sqrt[3]{x^3.y^3}=3xy\)

"="<=>x=y=1

Xét x,y\(< 0\)

\(\Rightarrow x^2-xy+y^2-x-y+1>0\)

\(\Rightarrow x+y=-1\)(vô lí vì x,yEZ)