National Olympiad Final Round

$$(x+y+z{)}^3=x^3+y^3+z^3+3(x^{2}y+x{y}^2+x^{2}z+y^{2}z+y^{2}x+xyz)+6xyz,$$ while $$\begin{gathered} (3(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)xyz=3(x+y+z)(xy+xz+yz) \\ =3(x^{2}y+x{y}^2+x^{2}z+x{z}^2+y^{2}z+y{z}^2)+9xyz. \end{gathered}$$ Thus $$(x+y+z{)}^3-3(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)xyz=x^3+y^3+z^3-3xyz.$$ By assumptions, $x+y+z=4$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3}$. Hence $64-4xyz=x^3+y^3+z^3-3xyz$, implying $x^3+y^3+z^3+xyz=64$. Consequently, the expression $x^3+y^3+z^3+xyz$ has only one value 64.