Question

Let the given polynomial be $P(x) = x^4 - 3x + 9$. The roots of $P(x)=0$ are $\alpha, \beta, \gamma, \delta$. Find the value of $(\alpha^3 + \beta^3 + \gamma^3 + \delta^3)$.

Answer 1

9

Answer 2

The equation is $x^4 + 0x^3 + 0x^2 - 3x + 9 = 0$. From Vieta's formulas: $e_1 = \sum \alpha = 0$ $e_2 = \sum \alpha\beta = 0$ $e_3 = \sum \alpha\beta\gamma = -(-3)/1 = 3$ $e_4 = \alpha\beta\gamma\delta = 9$ Using Newton's sums: $p_1 = e_1 = 0$ $p_2 = e_1 p_1 - 2e_2 = 0(0) - 2(0) = 0$ $p_3 = e_1 p_2 - e_2 p_1 + 3e_3 = 0(0) - 0(0) + 3(3) = 9$ Thus, $\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = 9$.