Thay x+y+z=1 vào biểu thức C, ta được:
\(C=\left(x+y+z-x\right)\left(x+y+z-y\right)\left(x+y+z-z\right)\)
\(C=\left(y+z\right)\left(z+x\right)\left(x+y\right)=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Ta có: \(x^3+y^3+z^3=\frac{1}{9}\Leftrightarrow\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{1}{9}\)
Thay x+y+z=1. Suy ra \(1-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{1}{9}\)
\(\Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{9}\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{9.3}=\frac{8}{27}\)
\(\Rightarrow C=\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{27}.\)
ĐS:...