If x = 332, y = 333, z = 335, then the value of x3 + y3 + z3 - 3xyz is?

Answer: Option A

Here, x = 332, y = 333, z = 335
Find x³ + y³ + z³ - 3xyz
$$=\frac{1}{2}\left(x+y+z\right)$$   $$\left[{\left(x-y\right)}^2+{\left(y-z\right)}^2+{\left(z-x\right)}^2\right]$$
$$=\left(\frac{332+333+335}{2}\right)$$   $$\left[{\left(332-333\right)}^2+{\left(333-335\right)}^2+{\left(332-335\right)}^2\right]$$
$$\begin{array}{rl}& =\frac{1000}{2}\left[1^2+2^2+3^2\right]\\ & =\frac{1000}{2}\left(14\right)\\ & =7000\end{array}$$