Question

M.n giải giúp em câu 5 với ạ

Answer 1

Vì x+y+z =1 nên \(x^3+y^3+x^3-3xyz=x^2+y^2+z^2-xy-yz-zx\)
\(=\dfrac{1}{2}\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]\)
\(=\dfrac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)
Vậy \(x^3+y^3+z^3-3xyz=\dfrac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\) (đpcm)