Ta có: \(a=x^3-3x^2+5x\)
\(< =>a=\left(x^3-3x^2+3x-1\right)+2x+1\)
\(< =>a=\left(x-1\right)^3+2x+1\)
Tương tự: \(b=\left(y-1\right)^3+2y+1\)
Do đó: \(a+b=\left(x-1\right)^3+\left(y-1\right)^3+2x+2y+2=6\)
\(< =>\left(x-1\right)^3+\left(y-1\right)^3+2x+2y-4=0\)
\(< =>\left(x-1\right)^3+\left(y-1\right)^3+2.\left(x-1\right)+2.\left(y-1\right)=0\)
Đặt x-1=c, y-1=d
\(=>c^3+d^3+2c+2d=0\)
\(< =>\left(c+d\right).\left(c^2-cd+d^2\right)+2\left(c+d\right)=0\)
\(< =>\left(c+d\right).\left(c^2-cd+d^2+2\right)=0\)
Vì \(c^2-cd+d^2+2>0< =>c^2-cd+d^2+2\ne0\)
<=>c+d=0
<=>x-1+y-1=0
<=>x+y=2
Vậy x+y=2
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