$x^{3} + px^{2} + qx + r = 0$, roots $\alpha, \beta, \gamma$. (a) Vieta; (b) $p_{2} = p^{2} - 2q$; (c) $\sum \alpha^{2}\beta^{2} = q^{2} - 2pr$; (d) $\sum 1/\alpha = -q/r$; (e) $p_{3}$ by Newton; (f) apply to $x^{3} - 4x^{2} + 5x - 2 = 0$.$x^{3} + px^{2} + qx + r = 0$，根 $\alpha, \beta, \gamma$。(a) 韦达；(b) $p_{2} = p^{2} - 2q$；(c) $\sum \alpha^{2}\beta^{2} = q^{2} - 2pr$；(d) $\sum 1/\alpha = -q/r$；(e) 用 Newton 求 $p_{3}$；(f) 应用于 $x^{3} - 4x^{2} + 5x - 2 = 0$。
Answers:答案： (a) $e_{1} = -p,\; e_{2} = q,\; e_{3} = -r$ · (e) $p_{3} = -p^{3} + 3pq - 3r$ · (f) $p_{2} = 6,\; \sum 1/\alpha = \tfrac{5}{2},\; p_{3} = 10$
(a) Vieta for the monic cubic A1·A1·A1
Expand $(x - \alpha)(x - \beta)(x - \gamma) = x^{3} - (\alpha + \beta + \gamma)x^{2} + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma$. Match to $x^{3} + px^{2} + qx + r$:
$$ e_{1} := \alpha + \beta + \gamma \;=\; -p, \qquad e_{2} := \alpha\beta + \beta\gamma + \gamma\alpha \;=\; q, \qquad e_{3} := \alpha\beta\gamma \;=\; -r. $$
(b) $p_{2} = p^{2} - 2q$ M1·A1
Square $e_{1}$:
$$ e_{1}^{2} \;=\; (\alpha + \beta + \gamma)^{2} \;=\; \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha\beta + \beta\gamma + \gamma\alpha) \;=\; p_{2} + 2 e_{2}. $$
Rearrange: $p_{2} = e_{1}^{2} - 2 e_{2} = (-p)^{2} - 2 q = p^{2} - 2q. \quad \text{AG}$
(c) $\alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2} = q^{2} - 2pr$ M1·A1
Square $e_{2}$:
$$ e_{2}^{2} \;=\; (\alpha\beta + \beta\gamma + \gamma\alpha)^{2} \;=\; \alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2} + 2(\alpha^{2}\beta\gamma + \alpha\beta^{2}\gamma + \alpha\beta\gamma^{2}). $$
The cross terms factor: $\alpha^{2}\beta\gamma + \alpha\beta^{2}\gamma + \alpha\beta\gamma^{2} = \alpha\beta\gamma(\alpha + \beta + \gamma) = e_{3}\,e_{1}$. So
$$ e_{2}^{2} \;=\; \sum \alpha^{2}\beta^{2} + 2 e_{1} e_{3}, \quad \Longrightarrow \quad \sum \alpha^{2}\beta^{2} \;=\; e_{2}^{2} - 2 e_{1} e_{3} \;=\; q^{2} - 2(-p)(-r) \;=\; q^{2} - 2pr. \quad \text{AG} $$
(d) $\sum 1/\alpha = -q/r$ M1·A1
Common denominator:
$$ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \;=\; \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \;=\; \frac{e_{2}}{e_{3}} \;=\; \frac{q}{-r} \;=\; -\frac{q}{r}. \quad \text{AG} $$
Non-degeneracy: requires $r \ne 0$ (i.e. $0$ is not a root of the cubic, so $\alpha, \beta, \gamma$ are all non-zero and the reciprocals exist).
(e) Newton's identity for $p_{3}$ M1·A1·A1
Newton: $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$, with $p_{1} = e_{1}$.
$$ p_{3} \;=\; e_{1}(e_{1}^{2} - 2 e_{2}) - e_{2} e_{1} + 3 e_{3} \;=\; e_{1}^{3} - 3 e_{1} e_{2} + 3 e_{3}. $$
Substitute Vieta:
$$ p_{3} \;=\; (-p)^{3} - 3(-p)(q) + 3(-r) \;=\; -p^{3} + 3 p q - 3 r. $$
(f) Apply to $x^{3} - 4 x^{2} + 5 x - 2 = 0$ A1·A1·A1
Here $p = -4$, $q = 5$, $r = -2$. Apply (b), (d), (e):
- $p_{2} = p^{2} - 2 q = 16 - 10 = 6$.
- $\sum 1/\alpha = -q/r = -5/(-2) = \tfrac{5}{2}$.
- $p_{3} = -p^{3} + 3 p q - 3 r = -(-64) + 3(-4)(5) - 3(-2) = 64 - 60 + 6 = 10$.
Verification by factoring. $x = 1$: $1 - 4 + 5 - 2 = 0$. $\checkmark$ Synthetic divide $1, -4, 5, -2 \;|\; 1$: $1, -3, 2, 0$. Quotient $x^{2} - 3x + 2 = (x - 1)(x - 2)$. So roots are $\alpha, \beta, \gamma = 1, 1, 2$ (double root at $1$, simple at $2$). Direct sums: $p_{2} = 1 + 1 + 4 = 6 \;\checkmark$; $\sum 1/\alpha = 1 + 1 + \tfrac{1}{2} = \tfrac{5}{2} \;\checkmark$; $p_{3} = 1 + 1 + 8 = 10 \;\checkmark$.
Newton's identities are the canonical machinery for power sums. For a degree-$n$ monic polynomial with elementary symmetric polynomials $e_{1}, \ldots, e_{n}$ and power sums $p_{k} = \sum \alpha_{i}^{k}$, Newton's identities give a recursion:
$$ p_{k} \;=\; e_{1} p_{k - 1} - e_{2} p_{k - 2} + \cdots + (-1)^{k - 1} k\, e_{k} \quad (k \le n), $$
with the convention $p_{0} = n$. For a cubic ($n = 3$): $p_{1} = e_{1}$; $p_{2} = e_{1} p_{1} - 2 e_{2}$; $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$; thereafter the recursion is purely $p_{k} = e_{1} p_{k - 1} - e_{2} p_{k - 2} + e_{3} p_{k - 3}$ for $k \ge 4$ (no new $e$'s, just the three you have). This is exactly the kind of structural calculation Paper 3 rewards: state the identity, plug Vieta, verify on a concrete cubic. The double root in (f) is a deliberate trap — students who write $\alpha = 1, \beta = 2, \gamma = ?$ and forget that the cubic has $1$ as a repeated root undercount $p_{2}$ and $p_{3}$. Newton's identities handle multiplicities automatically because Vieta itself counts roots with multiplicity.
(a) 首一三次的韦达 A1·A1·A1
展开 $(x - \alpha)(x - \beta)(x - \gamma) = x^{3} - (\alpha + \beta + \gamma)x^{2} + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma$，与 $x^{3} + px^{2} + qx + r$ 比对：
$$ e_{1} := \alpha + \beta + \gamma \;=\; -p, \qquad e_{2} := \alpha\beta + \beta\gamma + \gamma\alpha \;=\; q, \qquad e_{3} := \alpha\beta\gamma \;=\; -r. $$
(b) $p_{2} = p^{2} - 2q$ M1·A1
对 $e_{1}$ 平方：
$$ e_{1}^{2} \;=\; (\alpha + \beta + \gamma)^{2} \;=\; \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha\beta + \beta\gamma + \gamma\alpha) \;=\; p_{2} + 2 e_{2}. $$
移项：$p_{2} = e_{1}^{2} - 2 e_{2} = (-p)^{2} - 2 q = p^{2} - 2q$。$\quad\text{AG}$
(c) $\alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2} = q^{2} - 2pr$ M1·A1
对 $e_{2}$ 平方：
$$ e_{2}^{2} \;=\; \alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2} + 2(\alpha^{2}\beta\gamma + \alpha\beta^{2}\gamma + \alpha\beta\gamma^{2}). $$
交叉项可分解：$\alpha^{2}\beta\gamma + \alpha\beta^{2}\gamma + \alpha\beta\gamma^{2} = \alpha\beta\gamma(\alpha + \beta + \gamma) = e_{3}\,e_{1}$。故
$$ \sum \alpha^{2}\beta^{2} \;=\; e_{2}^{2} - 2 e_{1} e_{3} \;=\; q^{2} - 2(-p)(-r) \;=\; q^{2} - 2pr. \quad \text{AG} $$
(d) $\sum 1/\alpha = -q/r$ M1·A1
通分：
$$ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \;=\; \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \;=\; \frac{e_{2}}{e_{3}} \;=\; \frac{q}{-r} \;=\; -\frac{q}{r}. \quad \text{AG} $$
非退化条件：需 $r \ne 0$（即 $0$ 不是三次的根，$\alpha, \beta, \gamma$ 非零，倒数存在）。
(e) Newton 恒等式求 $p_{3}$ M1·A1·A1
Newton：$p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$，其中 $p_{1} = e_{1}$。
$$ p_{3} \;=\; e_{1}(e_{1}^{2} - 2 e_{2}) - e_{2} e_{1} + 3 e_{3} \;=\; e_{1}^{3} - 3 e_{1} e_{2} + 3 e_{3}. $$
代入韦达：
$$ p_{3} \;=\; (-p)^{3} - 3(-p)(q) + 3(-r) \;=\; -p^{3} + 3 p q - 3 r. $$
(f) 应用于 $x^{3} - 4 x^{2} + 5 x - 2 = 0$ A1·A1·A1
此处 $p = -4$、$q = 5$、$r = -2$。应用 (b)、(d)、(e)：
- $p_{2} = p^{2} - 2 q = 16 - 10 = 6$。
- $\sum 1/\alpha = -q/r = -5/(-2) = \tfrac{5}{2}$。
- $p_{3} = -p^{3} + 3 p q - 3 r = 64 - 60 + 6 = 10$。
因式验证。$x = 1$：$1 - 4 + 5 - 2 = 0$ $\checkmark$。综合除法 $1, -4, 5, -2 \mid 1$ 得 $1, -3, 2, 0$。商 $x^{2} - 3x + 2 = (x - 1)(x - 2)$。故根 $\alpha, \beta, \gamma = 1, 1, 2$（$1$ 为二重根，$2$ 简单根）。直接求和：$p_{2} = 1 + 1 + 4 = 6$ $\checkmark$；$\sum 1/\alpha = 1 + 1 + \tfrac{1}{2} = \tfrac{5}{2}$ $\checkmark$；$p_{3} = 1 + 1 + 8 = 10$ $\checkmark$。
Newton 恒等式是幂和的标准工具。$n$ 次首一多项式有初等对称多项式 $e_{1}, \ldots, e_{n}$ 与幂和 $p_{k} = \sum \alpha_{i}^{k}$，Newton 恒等式给出递推：
$$ p_{k} \;=\; e_{1} p_{k - 1} - e_{2} p_{k - 2} + \cdots + (-1)^{k - 1} k\, e_{k} \quad (k \le n), $$
约定 $p_{0} = n$。三次（$n = 3$）情形：$p_{1} = e_{1}$；$p_{2} = e_{1} p_{1} - 2 e_{2}$；$p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$；$k \ge 4$ 之后递推纯为 $p_{k} = e_{1} p_{k - 1} - e_{2} p_{k - 2} + e_{3} p_{k - 3}$（不再引入新的 $e$）。这正是 Paper 3 看重的结构化计算：写出恒等式、代入韦达、对具体三次核验。(f) 的二重根是故意设的陷阱——把根写成 $\alpha = 1, \beta = 2, \gamma = ?$ 而忘记 $1$ 是重根会把 $p_{2}$、$p_{3}$ 算少。Newton 恒等式因为韦达本身就按重数计数，自动处理重根。