Có :
x3 + y3 + z3 = 3xyz
x3 + y3 + z3 - 3xyz = 0
(x + y)3 - 3.xy.(x + y) + z3 - 3xyz = 0
(x + y)3 + z3 - 3xy.(x + y + z) = 0
(x + y + z).[(x + y)2 - (x + y).z) + z2] - 3xy(x + y + z) = 0
(x + y + z).[x2 + 2xy + y2 - zx - yz + z2] - 3xy(x + y + z) = 0
(x + y + z).[x2 + y2 + z2 - xy - yz - zx] = 0
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
Với \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)
a, x^4 - 5x^2 + 4
= x^4 - 4x^2- x+ 4
= x^2 . (x^2 - 4) - (x^2 - 4)
= (x^2 - 4) . (x^2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
