If x = 222, y = 223 and z = 224, then find the value of $x^3 + y^3 + z^3 - 3xyz$.

2007

x = 222,

y = 223

z = 224 

x³ + y³ + z³ - 3xyz = \(\frac{1}{2}\)(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

= x³ + y³ + z³ - 3xyz = \(\frac{1}{2}\)(222 + 223 + 224)[(222 - 223)² + (223 - 224)² + (224 - 222)²]

= x³ + y³ + z³ - 3xyz = \(\frac{1}{2}\)(669)[(1) + (1) + (4)]

= x³ + y³ + z³ - 3xyz = 3 × 669 

= x³ + y³ + z³ - 3xyz = 2007