+87 Given that \((x+y+z)(xy+xz+yz)=18\) and that \(x^2(y+z)+y^2(x+z)+z^2(x+y)=6\) for real numbers \(x\) , \(y\) , and \(z\), what is the value of \(xyz\)?
+129852 x^2 ( y + z) + y^2 ( x + z) + z^2 ( x + y) = 6
(x^2y + x^2z + xy^2 + zy^2 + xz^2 + yz^2) = 6
(x + y + z)(xy + xz + yz) = 18
x^2y + x^2z + xyz + xy^2 + xyz + y^2z + xyz + xz^2 + yz^2 = 18
(x^2y + x^2z + xy^2 + zy^2 + xz^2 + yz^2) + 3xyz = 18
(6) + 3xyz = 18
3xyz = 12
xyz = 4
