Let , , and be the roots of the cubic equation 2x^3 - 6x^2 + x + 5 = 0 . If ^2 + ^2 + ^2 = 12 , then the value of ^3 + ^3 + ^3 is equal to

Step-by-step solution

Let the roots of the equation 2x^3 - 6x^2 + x + 5 = 0 be , , . From Vieta's formulas, we have: + + = - -6 2 = 3 + + = 2 = - 5 2 We are given that ^2 + ^2 + ^2 = 12 . Using the identity ( + + )^2 = ^2 + ^2 + ^2 + 2( + + ) , we get: 3^2 = 12 + 2 ( 2 ) 9 = 12 + = -3 Thus, the sum of pairwise products is + + = - 3 2 . Now, using the identity for the sum of cubes: ^3 + ^3 + ^3 - 3 = ( + + )( ^2 + ^2 + ^2 - ( + + )) Substitute the known values: ^3 + ^3 + ^3 - 3 (- 5 2 ) = 3 (12 - (- 3 2 ) ) ^3 + ^3 + ^3 + 15 2 = 3 (12 +