Question -
Answer -
a + b + c = 9, ab + be + ca = 26
Squaring, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒ a2 + b2 + c2 + 2 x 26 = 81
⇒ a2 + b2 + c2 + 52 = 81
∴ a2 + b2 + c2 = 81 – 52 = 29
Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]
= 9[29 – 26]
= 9 x 3 = 27