If x = 255, y = 256, z = 257, then find the value of $$x^3 + y^3 + z^3 — 3xyz.$$

Given,  $$x$$ = 255,  $$y$$ = 256,  $$z$$ = 257

$$x^3+y^3+z^3—3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$$

$$=\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)$$

$$=\frac{\left(x+y+z\right)}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]$$

$$=\frac{\left(255+256+257\right)}{2}\left[\left(255-256\right)^2+\left(256-257\right)^2+\left(257-255\right)^2\right]$$

$$=\frac{768}{2}\left[1+1+4\right]$$

$$=\frac{768}{2}\left[6\right]$$

$$=2304$$

Hence, the correct answer is Option D