1. **State the problem:** We have a quartic equation $$x^4 + bx^3 + cx^2 + dx - 2 = 0$$ with roots $$\alpha, \beta, \gamma, \delta$$. Given: $$\alpha + \beta + \gamma + \delta = 3,$$ $$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 5,$$ $$\alpha^{-1} + \beta^{-1} + \gamma^{-1} + \delta^{-1} = 6.$$ (a) Find $$b, c, d$$. (b) Given also $$\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = -27$$, find $$\alpha^4 + \beta^4 + \gamma^4 + \delta^4$$. --- 2. **Recall Viète's formulas for quartic:** For roots $$\alpha, \beta, \gamma, \delta$$ of $$x^4 + bx^3 + cx^2 + dx + e = 0,$$ we have: - Sum of roots: $$\alpha + \beta + \gamma + \delta = -b$$ - Sum of products of roots two at a time: $$\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = c$$ - Sum of products of roots three at a time: $$\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -d$$ - Product of roots: $$\alpha\beta\gamma\delta = e$$ Given $$e = -2$$. 3. **Use given sums:** - From sum of roots: $$\alpha + \beta + \gamma + \delta = 3 = -b \implies b = -3.$$ 4. **Find sum of products of roots two at a time, $$c$$:** Use the identity: $$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta).$$ Plug in values: $$5 = 3^2 - 2c \implies 5 = 9 - 2c \implies 2c = 9 - 5 = 4 \implies c = 2.$$ 5. **Find $$d$$ using the sum of reciprocals:** Sum of reciprocals: $$\alpha^{-1} + \beta^{-1} + \gamma^{-1} + \delta^{-1} = 6.$$ Recall: $$\alpha^{-1} + \beta^{-1} + \gamma^{-1} + \delta^{-1} = \frac{\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta}{\alpha\beta\gamma\delta} = \frac{-d}{-2} = \frac{d}{2}.$$ So: $$6 = \frac{d}{2} \implies d = 12.$$ 6. **Summary for (a):** $$b = -3, \quad c = 2, \quad d = 12.$$ --- 7. **For (b), find $$\alpha^4 + \beta^4 + \gamma^4 + \delta^4$$ given $$\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = -27$$.** Use Newton's identities relating power sums $$p_k = \alpha^k + \beta^k + \gamma^k + \delta^k$$ to coefficients: For quartic $$x^4 + bx^3 + cx^2 + dx + e = 0$$ with roots $$r_i$$, Newton's sums say: $$p_1 + b = 0,$$ $$p_2 + b p_1 + 2 c = 0,$$ $$p_3 + b p_2 + c p_1 + 3 d = 0,$$ $$p_4 + b p_3 + c p_2 + d p_1 + 4 e = 0.$$ We know: - $$p_1 = 3$$ - $$p_2 = 5$$ - $$p_3 = -27$$ - $$b = -3, c = 2, d = 12, e = -2$$ Check first three for consistency: - $$p_1 + b = 3 - 3 = 0$$ correct. - $$p_2 + b p_1 + 2 c = 5 + (-3)(3) + 4 = 5 - 9 + 4 = 0$$ correct. - $$p_3 + b p_2 + c p_1 + 3 d = -27 + (-3)(5) + 2(3) + 36 = -27 - 15 + 6 + 36 = 0$$ correct. Now find $$p_4$$: $$p_4 + b p_3 + c p_2 + d p_1 + 4 e = 0$$ $$p_4 + (-3)(-27) + 2(5) + 12(3) + 4(-2) = 0$$ $$p_4 + 81 + 10 + 36 - 8 = 0$$ $$p_4 + 119 = 0$$ $$p_4 = -119.$$ --- **Final answers:** (a) $$b = -3, c = 2, d = 12$$ (b) $$\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = -119.$$