Question

Giải phương trình \(x^3+y^3-6xy+8=0\)

Answer 1

PT <=> \(\left(x+y\right)^3+8-3x^2y-3xy^2-6xy=0\)

\(\left(x+y+2\right)\left(x^2+2xy+y^2-2x-2y+4\right)-3xy\left(x+y+2\right)=0\)

<=> \(\left(x+y+2\right)\left(x^2-xy+y^2-2x-2y+4\right)=0\)

\(\left(x+y+2\right)\left[\frac{1}{2}\left(x^2-2xy+y^2\right)+\frac{1}{2}\left(x^2-4x+4\right)+\frac{1}{2}\left(y^2-4y+4\right)=0\right]\)

<=> \(\frac{1}{2}\left(x+y+2\right)\left[\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\right]=0\)

<=> x = y = 2