Question

(a^3+b^3+c^3 -3abc) ÷ (a+b+c)​

Answer 1

$\dfrac{ {a}^{3} + {b}^{3} + {c}^{3} - 3abc }{(a + b + c)}$

Identity to use : a³ + b³ + c³ - 3abc = (a+b+c)(a²+b²+c² - ab - bc - ca)

$\dfrac{(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }{a + b + c}$

(a+b+c) get's cancelled on the numerator and the denominator

Answer =≥ (a² + b² + c² - ab -bc - ca)

----------------------------------------------------------------------------------------------------------------------------------------------

Some more identities :

(a+b)² = a² + 2ab + b²

(a-b)² = a² - 2ab + b²

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

(x+a)(x+b) = x² + x(a+b) + ab

(a+b)³ = a³ + 3a²b + 3ab² + b³

a³ + b³ = (a+b)(a²-ab + b²)

(a-b)³ = a³ - 3a²b + 3ab² - b³

a³-b³ = (a-b)(a²+ab+b²)

Conditional identity:

if a+b+c = 0,

a³ + b³ + c³ = 3abc

____________________________________________________________________________