Question

If a + b + c = 9 and ab + bc + ca = 26, find the value of a³ + b³ + c³ – 3abc.

Answer 1

a + b + c = 9, ab + bc + ca = 26 

Squaring, a + b + c = 9 both sides, we get 

(a + b + c)² = (9)² 

a² + b² + c² + 2(ab + bc + ca) = 81

a² + b² + c² + 2 x 26 = 81 

a² + b² + c² + 52 = 81 

a² + b² + c² = 29 

Now, a³ + b³ + c³ – 3abc = (a + b + c) [(a² + b² + c² – (ab + bc + ca)] 

= 9[29 – 26] 

= 9 x 3 

= 27 

⇒ a³ + b³ + c³ – 3abc = 27