[กิตติ]

ผมคิดได้ $-12765$
$(x-\alpha )(x-\beta )(x-\gamma ) = x^3 -3x^2+2556x-2553=f(x)=0$
$(\alpha-1 )(\beta-1 )(\gamma-1 ) = \quad -\left\{\,(1-\alpha )(1-\beta )(1-\gamma )\right\} \quad \quad = -f(1)\quad \quad = -1\quad $
จากสมการ เราได้ว่า$ \alpha+\beta+\gamma = 3$
$(\alpha-1 )+(\beta-1 )+(\gamma-1 ) = 0$
$[(\alpha-1 )+(\beta-1 )+(\gamma-1 )]^2 =(\alpha-1 )^2+(\beta-1 )^2+(\gamma-1 )^2 +2\left\{\,(\alpha-1 )(\beta-1 )+(\alpha-1 )(\gamma-1 )+(\beta-1 )(\gamma-1 )\right\} $
$ (\alpha+\beta+\gamma)^2 = 9$
$\alpha^2+\beta^2+\gamma^2 +2(\alpha\beta+\alpha\beta+\beta\gamma)= 9$
$\alpha\beta+\alpha\beta+\beta\gamma = 2556$
$\alpha^2+\beta^2+\gamma^2 = -5103$
$\alpha^2+\beta^2+\gamma^2-2(\alpha+\beta+\gamma)+3 = -5106$
$(\alpha-1 )^2+(\beta-1 )^2+(\gamma-1 )^2 = -5106$
$\left\{\,(\alpha-1 )(\beta-1 )+(\alpha-1 )(\gamma-1 )+(\beta-1 )(\gamma-1 )\right\} = 2553$
จาก $(a+b+c)^3 = (a^3+b^3+c^3) + 3(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2) + 6abc$
$(a+b+c)^3 = (a^3+b^3+c^3) +3abc(a+b+c)(\frac{1}{a} +\frac{1}{b}+\frac{1}{c}) -3abc$
ถ้า$a+b+c=0$ สมการจะเหลือเป็น $(a+b+c)^3 = (a^3+b^3+c^3) -3abc$
ดังนั้น $(\alpha-1 )^3+(\beta-1 )^3+(\gamma-1 )^3 = -3$
$(a+b+c)^5 =(a+b+c)^3(a+b+c)^2$
$=\left\{\,(a^3+b^3+c^3)+3\right\} \left\{\,(a^2+b^2+c^2)+2(2553)\right\} $
$=(a^5+b^5+c^5)+[(abc)^2\left\{\,(a+b+c)(\frac{1}{a^2} +\frac{1}{b^2}+\frac{1}{c^2})-(\frac{1}{a} +\frac{1}{b}+\frac{1}{c}) \right\} ]+3(-5106)+2(-3)(2553)+2(3)(2553)$
เนื่องจาก $a+b+c=0$
$0= (a^5+b^5+c^5)-[(abc)^2\left\{\,(\frac{1}{a} +\frac{1}{b}+\frac{1}{c}) \right\} ]+3(-5106)$
$0= (a^5+b^5+c^5)-[(abc)\left\{\,ab +bc+ac\right\} ]+3(-5106)$
$0= (a^5+b^5+c^5)+\left\{\,ab +bc+ac\right\}+3(-5106)$
$(\alpha-1 )^5+(\beta-1 )^5+(\gamma-1 )^5 = 3(5106)-\left\{\,(\alpha-1 )(\beta-1 )+(\alpha-1 )(\gamma-1 )+(\beta-1 )(\gamma-1 )\right\}$
$=2(3)(2553)-2553 =5\times 2553 = 12765$

โจทย์ถาม$\dfrac{(\alpha-1 )^5+(\beta-1 )^5+(\gamma-1 )^5}{(\alpha-1 )(\beta-1 )(\gamma-1 ) } = -12765$
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