三次方程 $5x^3-2x+1=0$ 的根为 $\alpha$、$\beta$、$\gamma$。

(a) 求一个以 $\dfrac{1}{\alpha^2}$、$\dfrac{1}{\beta^2}$、$\dfrac{1}{\gamma^2}$ 为根的三次方程。

(b) 求 $\dfrac{1}{\alpha^4}+\dfrac{1}{\beta^4}+\dfrac{1}{\gamma^4}$ 的值。

(c) 再求 $\dfrac{1}{\alpha^6}+\dfrac{1}{\beta^6}+\dfrac{1}{\gamma^6}$ 和 $\dfrac{1}{\alpha^8}+\dfrac{1}{\beta^8}+\dfrac{1}{\gamma^8}$ 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。

先把原根取倒数，再以其平方作为新根。用新变量的代数关系消去奇次方，避免对复根任意选择平方根分支。

The original roots are non-zero because the constant term is $1$. Put $z=1/x$. Dividing the original equation by $x^3$ gives

Now put $y=z^2=1/x^2$. Then $zy=2y-5$. Squaring gives

Thus the required cubic equation is

也可先把原方程整理成含x与x平方的乘积，平方后再直接代入y=1/x平方。

Rearranging the original equation,

Squaring and setting $y=1/x^2$,

Multiplication by $y^3$ gives $(5-2y)^2=y^3$. Hence the required equation is

把三个新根记作p、q、r，用韦达关系计算平方和。

Let $p=\alpha^{-2}$, $q=\beta^{-2}$, $r=\gamma^{-2}$. From the cubic in (a),

Therefore

每个新根都满足同一三次方程。先直接把三条方程相加求立方和，再乘根后相加求四次方和。

Write $S_j=p^j+q^j+r^j$. We have $S_1=4$, $S_2=-24$, and $S_0=3$.

Each root $y$ satisfies $y^3=4y^2-20y+25$. Summing over the three roots,

Multiplying the root equation by $y$ and summing gives

Hence

Also,