Question

If \( \alpha_1, \alpha_2, \alpha_3 \) are the roots of the equation \( x^3 + 3x + 2 = 0 \), then \( \alpha_1^5 + \alpha_2^5 + \alpha_3^5 \) is:

• A. \( -30 \)
• B. 6
• C. \( -6 \)
• D. 30

Hint:

To solve for higher powers of the roots, use relations from Vieta's formulas and simplify expressions by breaking them into lower powers.

Answer 1

The Correct Option is D

We are given the cubic equation: \[ x^3 + 3x + 2 = 0 \] By Vieta's formulas, the sum and product of the roots are: \[ \alpha_1 + \alpha_2 + \alpha_3 = 0, \quad \alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1 = 3, \quad \alpha_1\alpha_2\alpha_3 = -2 \] We need to find \( \alpha_1^5 + \alpha_2^5 + \alpha_3^5 \). Using the relationships from the original equation, we can express \( \alpha_1^5, \alpha_2^5, \alpha_3^5 \) in terms of lower powers and constants, and after simplifying, we find: \[ \alpha_1^5 + \alpha_2^5 + \alpha_3^5 = 30 \] Thus, the correct answer is 30.