If x + y + z = 6 and x2 + y2 + z2 = 20, then the value of x3 + y3 + z3 - 3xyz is?

Answer: Option C

$$\begin{array}{rl}& x+y+z=6\\ & x^2+y^2+z^2=20\\ & \Rightarrow {\left(x+y+z\right)}^2={\left(6\right)}^2\\ & \\ & \Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=36\\ & \\ & \Rightarrow 20+2\left(xy+yz+zx\right)=36\\ & \\ & \Rightarrow 2\left(xy+yz+zx\right)=16\\ & \\ & \Rightarrow xy+yz+zx=8\\ & \therefore x^3+y^3+z^3-3xyz\\ & =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-yz\right)\\ & =x^3+y^3+z^3-3xyz=6\left(20-8\right)\\ & =6\left(20-8\right)\\ & =6\times 12\\ & =72\end{array}$$