If $x + y + z = 19, xyz = 216$ and $xy + yz + zx = 114$, then the value of $ x^{3}+ y^{3}+z^{3} + xyz$ is:

1225

Given,

 x + y + z = 19

xy + yz + zx = 114

 (x + y + z)² = (19)²

x² + y² + z² + 2(xy + yz + zx) = 361

= x² + y² + z² = 361 − 2(xy + yz + zx)

= x² + y² + z² = 361 – 2 × 114 = 133

x² + y² + z² = 133

We know , x³ + y³ + z³ − 3xyz = (x + y + z)( x² + y² + z² − xy − yz − zx)

x³ + y³ + z³ − 3xyz = 19 × [133 – 114]

Adding 4xyz both sides,

x³ + y³ + z³ + xyz = 19 × 19 + 4 × 216 = 1225