# Quod Erat Demonstrandum

## 2013/10/24

### old question is so…

Filed under: Pure Mathematics — johnmayhk @ 4:23 下午

Question 1

If $\alpha,\beta,\gamma$ are roots of $x^3+px+q=0$, express

$(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2$

in terms of $p$ and $q$.

$\Delta=b^2-4ac$

$a=1$，我們有：

$\Delta=(\alpha-\beta)^2$

$\Delta=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2$

$\alpha,\beta,\gamma$$x^3+px+q=0$ 的根，故

$x^3+px+q\equiv (x-\alpha)(x-\beta)(x-\gamma)$

$\frac{d}{dx}(x^3+px+q)\equiv \frac{d}{dx}(x-\alpha)(x-\beta)(x-\gamma)$

$3x^2+p\equiv (x-\alpha)(x-\beta)+(x-\beta)(x-\gamma)+(x-\gamma)(x-\alpha)$

$3\alpha^2+p=(\alpha-\beta)(\alpha-\gamma)$
$3\beta^2+p=(\beta-\gamma)(\beta-\alpha)$
$3\gamma^2+p=(\gamma-\alpha)(\gamma-\beta)$

$(3\alpha^2+p)(3\beta^2+p)(3\gamma^2+p)=-(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2$

$(3\alpha^2+p)(3\beta^2+p)(3\gamma^2+p)$
$=p^3+3(\alpha^2+\beta^2+\gamma^2)p^2+9(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)p+27\alpha^2\beta^2\gamma^2$………………(*)

$x^3+px+q\equiv (x-\alpha)(x-\beta)(x-\gamma)$

$x^3+px+q\equiv x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma$

$\alpha+\beta+\gamma=0$
$\alpha\beta+\beta\gamma+\gamma\alpha=p$
$\alpha\beta\gamma=-q$

$(\alpha+\beta+\gamma)^2=0$
$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)=0$
$\alpha^2+\beta^2+\gamma^2=-2p$

$\alpha\beta+\beta\gamma+\gamma\alpha=p$
$(\alpha\beta+\beta\gamma+\gamma\alpha)^2=p^2$
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\beta\gamma^2)=p^2$
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)=p^2$
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=p^2$

$(3\alpha^2+p)(3\beta^2+p)(3\gamma^2+p)$
$=p^3+3(-2p)p^2+9(p^2)p+27q^2$
$=4p^3+27q^2$

$(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2=-(4p^3+27q^2)$

Question 2

If $\alpha,\beta,\gamma$ are real numbers such that $\alpha+\beta+\gamma=0$, prove that

$6(\alpha^3+\beta^3+\gamma^3)^2\le (\alpha^2+\beta^2+\gamma^2)^3$

$\alpha+\beta+\gamma=0$，我們不妨設

$\alpha,\beta,\gamma$ 為下式的實根（real roots）：

$x^3+px+q=0$

Question 1，知

$(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2=-(4p^3+27q^2)$

$\alpha,\beta,\gamma$ 為實數，即上式等號左面一定不少於零，故

$-(4p^3+27q^2)\ge 0$

$\Rightarrow 4p^3+27q^2\le 0$……………….(1)

Question 1 的做數過程中，有

$\alpha^2+\beta^2+\gamma^2=-2p\Rightarrow (\alpha^2+\beta^2+\gamma^2)^3=-8p^3$……………….(2)

$\alpha^3+p\alpha+q=0$
$\beta^3+p\beta+q=0$
$\gamma^3+p\gamma+q=0$

$(\alpha^3+\beta^3+\gamma^3)+p(\alpha+\beta+\gamma)+3q=0$

$(\alpha^3+\beta^3+\gamma^3)=-3q\Rightarrow (\alpha^3+\beta^3+\gamma^3)^2=9q^2$……………….(3)

$6(\alpha^3+\beta^3+\gamma^3)^2\le (\alpha^2+\beta^2+\gamma^2)^3$