四次方程 $x^4+2x^3-1=0$ 的根为 $\alpha$、$\beta$、$\gamma$、$\delta$。

(a) 求一个以 $\alpha^4$、$\beta^4$、$\gamma^4$、$\delta^4$ 为根的四次方程，并写出 $\alpha^4+\beta^4+\gamma^4+\delta^4$ 的值。

(b) 求 $\alpha^5+\beta^5+\gamma^5+\delta^5$ 的值。

(c) 求 $\alpha^8+\beta^8+\gamma^8+\delta^8$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

令新根y=x的四次方，原方程给出x的三次方与y的关系。取四次方消去x，再用韦达定理读取新根之和。

For any root $x$ of the original equation, put $y=x^4$. Then

Raising to the fourth power and using $x^{12}=y^3$ gives

Thus a required quartic equation is

Its roots are $\alpha^4,\beta^4,\gamma^4,\delta^4$, so Vieta’s formula gives

将每个根满足的原方程乘以该根，再把四个等式相加，就能把五次和降成四次和及一次和。

From the original quartic, $\alpha+\beta+\gamma+\delta=-2$.

Multiplying $x^4+2x^3-1=0$ by $x$ gives $x^5+2x^4-x=0$. Summing over the four roots and using (a),

八次方是新根的平方。用新四次方程的二次对称和与一次和，立即得到平方和。

By (a), the sum of the new roots is $20$ and their pairwise product sum is $6$. Hence

Here the pairwise sum contains the six products of two distinct new roots.