The value of $a^3 + b^3 + c^3 - 3abc$, when a = 125, b = 127 and c = 129, is :

4572

a = 125

 b = 127

c = 129

We know that,

a³ + b³ + c³ - 3abc = \(\frac{1}{2}\) × (a + b + c) × [(a - b)² + (b - c)² + (c - a)²]

According to the question,

= a³ + b³ + c³ - 3abc = \(\frac{1}{2}\) × (125 + 127 + 129) × [(125 - 127)² + (127 - 129)² + (125 - 129)²]

= a³ + b³ + c³ - 3abc = \(\frac{1}{2}\) × 381 × [4 + 4 + 16]

= a³ + b³ + c³ - 3abc =\(\frac{1}{2}\) × 381 × 24

= a³ + b³ + c³ - 3abc = 12 × 381

= a³ + b³ + c³ - 3abc = 4572