三次方程的根 等幂和

hbghlyj

Hall and Knight - Higher Algebra Volume 2 page 504:
140. If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+qx+r=0$, prove that
$$3\left(\alpha^{2}+\beta^2+\gamma^{2}\right)\left(\alpha^5+\beta^{5}+\gamma^{5}\right)=5\left(\alpha^{3}+\beta^{3}+\gamma^{3}\right)\left(\alpha^{4}+\beta^4+\gamma^{4}\right)$$

[St. John's College, Cambridge]

142. If $a, b, c$ are the roots of the equation $x^3+q x^2+r=0$, form the equation whose roots are $a+b-c, b+c-a, c+a-b$.

hbghlyj

140. 相关: 正文 Art. 526 (page 442)

Art. 526

In order to find the values of expressions involving $a, b, c$ when these quantities are connected by the equation $a+b+c=0$, we might employ the substitution
$$
a=h+k, \quad b=\omega h+\omega^2 k, \quad c=\omega^2 h+\omega k
$$
If however the expressions involve $a, b, c$ symmetrically the method exhibited in the following example is preferable.
Example. If $a+b+c=0$, shew thet
$$
6\left(a^5+b^5+c^5\right)=5\left(a^3+b^3+c^3\right)\left(a^2+b^2+c^2\right) \text {. }
$$
We have identically
\[(1+a x)(1+b x)(1+c x)=1+p x+q x^2+r x^3 \]
where\[p=a+b+c, q=b c+c a+a b, r=a b c\]
Hence, using the condition given,
$$
(1+a x)(1+b x)(1+c x)=1+q x^2+r x^3
$$
Taking logarithms and equating the coefficients of $x^n$, we have\begin{align*}&\frac{(-1)^{n-1}}{n}\left(a^n+b^n+c^n\right)\\&=\text{coefficient of }x^n\text{ in the expansion of }\log \left(1+q x^2+r x^3\right)\\&=\text{coefficient of $x^n$ in }\left(q x^2+r x^3\right)-\frac{1}{2}\left(q x^2+r x^3\right)^2+\frac{1}{3}\left(q x^2+r x^3\right)^3-\ldots\end{align*}
By putting $n=2,3,5$ we obtain$${a^2+b^2+c^2\over2}=q,{a^3+b^3+c^3\over3}=r,{a^5+b^5+c^5\over5}=-qr$$
whence$${a^{5}+b^{5}+c^{5} \over5}=\frac{a^3+ b^{3}+c^{3}}{3}\cdot\frac{a^{2}+b^{2}+c^{2}}{2}$$and the required result at once follows.
If $a=\beta-\gamma, b=\gamma-\alpha, c=\alpha-\beta$, the given condition is satisfied; hence we have identically for all values of $\alpha, \beta, \gamma$
$$
\begin{aligned}
&6\left\{(\beta-\gamma)^5+(\gamma-\alpha)^5+(\alpha-\beta)^5\right\} \\
& =5\left\{(\beta-\gamma)^3+(\gamma -\alpha)^3+(\alpha -\beta)^3\right\}\left\{(\beta-\gamma)^2+(\gamma-\alpha)^2+(\alpha-\beta)^2\right\}
\end{aligned}
$$
that is,
$$
(\beta-\gamma)^5+(\gamma-\alpha)^5+(\alpha-\beta)^5=5(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)\left(\alpha^2+\beta^2+\gamma^2-\beta \gamma-\gamma \alpha-\alpha \beta\right) ;
$$
compare Ex. 3, Art. 522.

对于140, 由$(x-\alpha)(x-\beta)(x-\gamma)\equiv x^3+qx+r$ 令$y=\frac1x$得
$$(1-\alpha y)(1-\beta y)(1-\gamma y)\equiv1+q y^2+ry^3$$取log$$\log(1-\alpha y)+\log(1-\beta y)+\log(1-\gamma y)\equiv\log(1+q y^2+ry^3)$$级数展开(右边只需写出前2项,因为第3项最低已经是6次)\begin{align*}\sum_{n=1}^\infty-\frac{\alpha^n+\beta^n+\gamma^n}ny^n&\equiv\log \left(1+q y^2+r y^3\right)\\&\equiv\left(q y^2+r y^3\right)-\frac{1}{2}\left(q y^2+r y^3\right)^2+\ldots\end{align*}对比$y^2,y^3,y^4,y^5$系数
\begin{matrix}\frac{\alpha^{2}+\beta^{2}+\gamma^{2}}{2}=-q,&\frac{\alpha^{3}+\beta^{3}+\gamma^{3}}{3}=-r,\\ \frac{\alpha^{4}+\beta^{4}+\gamma^{4}}{4}=\frac{q^{2}}{2}, &\frac{\alpha^{5}+\beta^{5}+\gamma^{5}}{5}=qr\end{matrix}因此$$\frac{\alpha^{2}+\beta^{2}+\gamma^{2}}{2}\frac{\alpha^{5}+\beta^{5}+\gamma^{5}}{5}=-q^2r=\frac{\alpha^{3}+\beta^{3}+\gamma^{3}}{3} \frac{\alpha^{4}+\beta^{4}+\gamma^{4}}{2}$$
即$$3\left(\alpha^{2}+\beta^2+\gamma^{2}\right)\left(\alpha^5+\beta^{5}+\gamma^{5}\right)=5\left(\alpha^{3}+\beta^{3}+\gamma^{3}\right)\left(\alpha^{4}+\beta^4+\gamma^{4}\right)$$

色k

你是否打算翻译？

hbghlyj

由$a+b+c=-q$得$$c=\frac{-q-(a+b-c)}2, a=\frac{-q-(b+c-a)}2, b=\frac{-q-(c+a-b)}2$$满足的方程为$x^3+q x^2+r=0$
所以$a+b-c, b+c-a, c+a-b$满足的方程为
$$\left(\frac{-q-x}2\right)^3+q\left(\frac{-q-x}2\right)^2+r=0$$

hbghlyj

例3.5 已知$x+y+z=0$, 求证$\frac{x^{5}+y^{5}+z^{5}}{5}=\frac{x^{3}+y^{3}+z^{3}}{3} \cdot \frac{x^{2}+y^{2}+z^{2}}{2}$