Question

Cho x+y+z=0. CMR x³ + y³ + z³ = 3xyz

Answer 1

Xét \(A=x^3+y^3+z^3-3xyz\)

\(=\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3x^2y-3xy^2-3xyz\)

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

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz+yz+z^2\right)-3xy\left(x+y+z\right)\)

Với \(x+y+z=0\) thì  \(A=0.\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy.0=0\)

\(A=x^3+y^3+z^3-3xyz=0\Rightarrow x^3+y^3+z^3=3xyz\) (đpcm)

Answer 2

x+y+z=0=> (x+y+z)(x²+y²+z²-xy-yz-zx)=0 (*)

Nhân (*) ra được :

x³+y³+z³-3xyz=0<=> x³+y³+z³= 3xyz(đpcm)