Given that (x+y+z)(xy+xz+yz)=25 and that x^3+y^3+z^3=18 for real numbers x, y, and z, what is the value of xyz?

Answer 1

We expand $(x+y+z)(xy+xz+yz)$, to find that $(x+y+z)(xy+xz+yz)=x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz=25$. Similarly, we have $(x+y+z)(xy+xz+yz)=xy^2+xyz+xyz+xz^2+xyz+xyz+yz^2+xyz+xyz=3xyz+xy^2+x^2z+xz^2+y^2z+yz^2=25$. Subtracting these two equations, we find that $2xyz=0$, so $\boxed{xyz = 0}$.