a³ + b³ + c³ Formula — Algebraic Identity

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

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

Special Case (most important for exams): If a + b + c = 0, then: a³ + b³ + c³ − 3abc = 0 ∴ a³ + b³ + c³ = 3abc

This special case appears frequently in competitive exams and Class 9 NCERT Polynomials.

Note on the second factor: a² + b² + c² − ab − bc − ca = ½[(a−b)² + (b−c)² + (c−a)²]

This means the second factor is always ≥ 0 for real values.

a³ + b³ = (a + b)(a² − ab + b²) a³ − b³ = (a − b)(a² + ab + b²)

(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b) (a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca → a² + b² + c² = (a+b+c)² − 2(ab+bc+ca)

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

If a = b = c: a³ + b³ + c³ − 3abc = 3a³ − 3a³ = 0 (consistent, since a+b+c = 3a ≠ 0 but a²+b²+c²−ab−bc−ca = 3a²−3a² = 0)

Example 1: Find a³+b³+c³ if a+b+c=0 and abc=5. Solution: Since a+b+c=0, a³+b³+c³ = 3abc = 3×5 = 15

Example 2: If a=1, b=2, c=−3, find a³+b³+c³−3abc. Check: a+b+c = 1+2+(−3) = 0 Since a+b+c = 0: a³+b³+c³ = 3abc = 3×(1)(2)(−3) = −18 So a³+b³+c³−3abc = −18−3(−6) = −18+18 = 0 ✓

Example 3: Find (2)³+(3)³+(−5)³ without direct calculation. Check: 2+3+(−5) = 0 So: 2³+3³+(−5)³ = 3×(2)(3)(−5) = 3×(−30) = −90 Verification: 8+27+(−125) = 35−125 = −90 ✓

Example 4: Factorise a³+b³+c³−3abc when a+b+c=5, a²+b²+c²=11, ab+bc+ca=7. Note: a²+b²+c²−ab−bc−ca = 11−7 = 4 a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca) = 5×4 = 20