If x + y + z = 17, xyz = 171 and xy + yz + zx = 111, then the value of $$\root 3 \of {\left( {{x^3} + {y^3} + {z^3} + xyz} \right)} $$     is:

A. -64

B. 0

C. 4

D. -4

Answer: Option D

Solution(By Examveda Team)

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
$$\eqalign{ & \root 3 \of {\left( {{x^3} + {y^3} + {z^3} + xyz} \right)} \cr & = \root 3 \of { - 64} \cr & = - 4 \cr} $$