If x + y + z = 17, xyz = 171 and xy + yz + zx = 111, then the value of 3 ( x^3 + y^3 + z^3 + xyz ) is:

Answer: Option D

x³ + y³ + z³ - 3xyz = (x + y + z)[x² + y² + z² - (xy + yz + zx)]
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
289 = x² + y² + z² + 2 × 111
x² + y² + z² = 67
Now,
x³ + y³ + z³ - 3xyz = 17(67 - 111)
x³ + y³ + z³ - 3xyz = -44 × 17
x³ + y³ + z³ - 3xyz + 4xyz = -748 + 4xyz
x³ + y³ + z³ + xyz = -748 + 4 × 171
x³ + y³ + z³ + xyz = -748 + 684
x³ + y³ + z³ + xyz = -64
$$\begin{array}{rl}& \text{root}3\text{of}\left(x^3+y^3+z^3+xyz\right)\\ & =\text{root}3\text{of}-64\\ & =-4\end{array}$$