If 1 α , 1 β are the roots of the equation a x 2 + b x + 1 = 0 , a ≠ 0 , a , b ∈ R , then the equation x x + b 3 + a 3 - 3 a b x = 0 has roots:

Step-by-step solution

a x 2 + b x + 1 = 0 has roots 1 α , 1 β ⇒    1 α + 1 β = - b a             . . . 1 1 α β = 1 a         . . . 2 x x + b 3 + a 3 - 3 a b x = 0 ⇒   x 2 + b 3 - 3 a b x + a 3 = 0 Multiply and divide with a 3 , we get ⇒       x 2 + a 3 b 3 a 3 - 3 1 a b a x + a 3 = 0 From equations 1   &   2 , ⇒    x 2 + α β 3 / 2 - α 1 / 2 + β 1 / 2 3