a³ + b³ + c³ − 3abc Formula and Proof

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

This formula expresses the difference between the sum of cubes and 3abc as the product of two polynomial factors.

A very common exam application of this formula happens when the sum of the three variables is zero.

If a + b + c = 0: Then the right side of the equation becomes: (0) × (a² + b² + c² − ab − bc − ca) = 0.

Therefore, the formula reduces to: a³ + b³ + c³ − 3abc = 0

Which means: a³ + b³ + c³ = 3abc

This is a powerful shortcut! If you need to find the sum of three cubes, and the base numbers add up to 0, their sum of cubes is simply 3 times their product.

Sometimes the formula is written using a half-factor to involve perfect squares:

a³ + b³ + c³ − 3abc = ½(a + b + c) [(a − b)² + (b − c)² + (c − a)²]

This form is particularly useful when proving inequalities because the sum of squares [(a−b)² + (b−c)² + (c−a)²] is always positive or zero.

Question: Without actually calculating the cubes, find the value of (12)³ + (−7)³ + (−5)³.

Solution: Let a = 12, b = −7, c = −5. First, check their sum: a + b + c = 12 + (−7) + (−5) = 12 − 12 = 0.

Since a + b + c = 0, we use the conditional identity: a³ + b³ + c³ = 3abc

(12)³ + (−7)³ + (−5)³ = 3 × (12) × (−7) × (−5) = 3 × 12 × 35 = 36 × 35 = 1260