If a + b + c = 11 and ab + bc + ca = 17, then what is the value of a3 + b3 + c3 – 3abc?

Given a3 + b3 + c3 – 3abccan be written as (a + b + c) {(a + b + c) 2 – 3(ab + bc + ca) }

⇒ a3 + b3 + c3 – 3abc= (a + b + c) {(a + b + c) 2 – 3(ab + bc + ca) } Using the given values

⇒ a3 + b3 + c3 – 3abc = 11 × (112 - 3 × 17) = 770

∴ a3 + b3 + c3 – 3abc = 770