If x = 999, y = 1000, z = 1001 then the value of x^3 + y^3 + z^3 - 3xyzx - y + z is?

Answer: Option D

$$\begin{array}{rl}& \therefore a^3+b^3+c^3-3abc\\ & =\frac{1}{2}\left(a+b+c\right)\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]\\ & \therefore \frac{x^3+y^3+z^3-3xyz}{x-y+z}\end{array}$$
$$=\frac{\frac{1}{2}\left(x+y+z\right)\left[{\left(x-y\right)}^2+{\left(y-z\right)}^2+{\left(z-x\right)}^2\right]}{x-y+z}$$
$$\begin{array}{rl}& =\frac{\frac{1}{2}\left(999+1000+1001\right)\left(1+1+4\right)}{999-1000+1001}\\ & =\frac{\frac{1}{2}\times 6\times 3000}{1000}\\ & =9\end{array}$$