Question

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

Answer 1

We have a + b + c = 9 ...(i) 

⇒ (a + b + c)² = 81 [On squaring both sides of (i)] 

⇒ a² + b² + c² + 2(ab + bc + ac) = 81 

⇒ a² + b² + c² + 2 × 26 = 81 [∵ab + bc + ac = 26] 

⇒ a² + b² + c² = (81 - 52) 

⇒ a² + b² + 2 = 29. 

Now, we have

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

 = (a + b + c) [(a² + b² + c² ) - (ab + bc + ac)] 

 = 9 × [(29 - 26)] 

 = (9 × 3) 

 = 27