If x + y + z = 19, xy + yz + zx = 144, then the value of x^3 + y^3 + z^3 - 3xyz is:

Answer: Option C

$$\begin{array}{rl}& x+y+z=19,\\ & xy+yz+zx=144,\\ & \sqrt{x^3+y^3+z^3-3xyz}\\ & \text{Let }z=0\\ & x+y=19,xy=144,\sqrt{x^3+y^3}=?\\ & \sqrt{x^3+y^3}\\ & =\sqrt{\left(x+y\right)\left[{\left(x+y\right)}^2-3xy\right]}\\ & =\sqrt{19\left({19}^2-3\times 144\right)}\\ & =\sqrt{19\times 19}\\ & =19\end{array}$$