If x = 255, y = 256, z = 257, then find the value of $x^3 + y^3 +z^3 -3xyz$.

2304

Given,

x = 255, y = 256 and z = 257

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

According to the question

= x³ + y³ + z³  - 3xyz = \(\frac{1}{2}\) × (255 + 256 + 257) × [(255 - 256)2 + (256 - 257)2 + (257 - 255)2]

= x³ + y³ + z³  - 3xyz = \(\frac{1}{2}\) × 768 × [(-1)² + (-1)² + (-2)²]

= x³ + y³ + z³  - 3xyz = \(\frac{1}{2}\) × 768 × [1 + 1 + 4]

= x³ + y³ + z³  - 3xyz = \(\frac{1}{2}\) × 768 × 6

= x³ + y³ + z³  - 3xyz = 768 × 3 = 2304