If A = a b c b c a c a b ,   a b c = 1 and A T A = I , then the value of a 3 + b 3 + c 3 (where a , b , c > 0 and A T is transpose of matrix A ) is

Step-by-step solution

A T A = I As A T = A ⇒ A 2 = I ⇒ a 2 + b 2 + c 2 = 1 and a b + b c + c a = 0 As a + b + c 2 = a 2 + b 2 + c 2 + 2 a b + b c + c a ⇒ a + b + c = 1 Now, a 3 + b 3 + c 3 - 3 a b c = a + b + c a 2 + b 2 + c 2 - a b - b c - c a ⇒ a 3 + b 3 + c 3 - 3 = a + b + c ⇒ a 3 + b 3 + c 3 = 3 + 1 ⇒ a 3 + b 3 + c 3 = 4