If x + y + z = 19, xy + yz + zx = 144, then the value of $$\sqrt {{x^3} + {y^3} + {z^3} - 3xyz} $$     is:

A. 21

B. 17

C. 19

D. 13

Answer: Option C

Solution(By Examveda Team)

$$\eqalign{ & x + y + z = 19, \cr & xy + yz + zx = 144, \cr & \sqrt {{x^3} + {y^3} + {z^3} - 3xyz} \cr & {\text{Let }}z = 0 \cr & x + y = 19,\,xy = 144,\,\sqrt {{x^3} + {y^3}} = ? \cr & \sqrt {{x^3} + {y^3}} \cr & = \sqrt {\left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} - 3xy} \right]} \cr & = \sqrt {19\left( {{{19}^2} - 3 \times 144} \right)} \cr & = \sqrt {19 \times 19} \cr & = 19 \cr} $$