檢視一元三次方程式的原始碼

[[分類:數學]]
==參數法簡表==
<table class=nicetable>
<tr><th rowspan=2>n</th><th rowspan=2>1,<br/>n,<br/><span style='font-size:80%'>-(n+1)</span></th><td rowspan=2>x<sup>3</sup>-3px+q=0<br/>設 x=u+v<br/><span style='font-size:80%'>或 x=uω<sup>2</sup>+vω<br/>或 x=uω+vω<sup>2</sup></span></td><td rowspan=2><span style='font-size:80%'>設 uv=p,u<sup>3</sup>+v<sup>3</sup>=-q<br/>則 [uω,vω<sup>2</sup>] 與<br/>[uω<sup>2</sup>,vω] 兩數對的<br/>乘積=p且立方和=-q</span></td><th rowspan=2><math>Z^2+n(n+1)Z+[\frac{n(n+1)+1}{3}]^3=0</math></th><th colspan=2>
<math>Z=\frac{-n(n+1)}{2} \pm \sqrt{[\frac{n(n+1)}{2}]^2-\frac{[n(n+1)+1]^3}{3^3}}</math>
</th><th rowspan=2>u,v=<math>\frac{n}{2}</math>±<math>\frac{n+2}{2\sqrt{3}}i</math></th></tr>
<tr><th colspan=2>
<math>Z=\frac{-n(n+1)}{2} \pm \frac{(n-1)(2n+1)(n+2)}{6\sqrt{3}}i</math>
</th></tr>

<tr><th>n=ω</th><th>1,ω,ω<sup>2</sup></th><th>x<sup>3</sup>-0x-1=0</th><th>uv=0,u<sup>3</sup>+v<sup>3</sup>=1</th><th>Z<sup>2</sup>-Z=0</th><th>Z=0,1</th><th>u,v=0,1 或0,ω 或0,ω<sup>2</sup></th><th>u,v=<math>\frac{ω}{2}</math>±<math>\frac{ω+2}{2\sqrt{3}}i</math></th></tr>

<tr><th>n=-1</th><th>1,-1,0</th><th>x<sup>3</sup>-x=0</th><th>uv=1/3,u<sup>3</sup>+v<sup>3</sup>=0</th><th>Z<sup>2</sup>+<math>\frac{1}{27}</math>=0</th><th>Z=±<math>\frac{1}{3\sqrt{3}}</math>i</th><th>u,v=±<math>\frac{1}{\sqrt{3}}</math>i</th><th>uω,vω<sup>2</sup>=<br/><math>\frac{-1}{2}</math>±<math>\frac{-1+2}{2\sqrt{3}}i</math></th></tr>

<tr><th>n=0 </th><th>1,0,-1</th><th>x<sup>3</sup>-x=0</th><th>uv=1/3,u<sup>3</sup>+v<sup>3</sup>=0</th><th>Z<sup>2</sup>+<math>\frac{1}{27}</math>=0</th><th>Z=±<math>\frac{1}{3\sqrt{3}}</math>i</th><th>u,v=±<math>\frac{1}{\sqrt{3}}</math>i</th><th>u,v=<math>\frac{0}{2}</math>±<math>\frac{0+2}{2\sqrt{3}}i</math></th></tr>

<tr><th>n=1 </th><th>1,1,-2</th><th>x<sup>3</sup>-3x+2=0</th><th>uv=1,u<sup>3</sup>+v<sup>3</sup>=-2</th><th>Z<sup>2</sup>+2Z+1=0</th><th>Z=-1,-1</th><th>u,v=-1,-1 或-ω,-ω<sup>2</sup></th><th>u,v=<math>\frac{1}{2}</math>±<math>\frac{1+2}{2\sqrt{3}}i</math></th></tr>

<tr><th>n=2 </th><th>1,2,-3</th><th>x<sup>3</sup>-7x+6=0</th><th>uv=7/3,u<sup>3</sup>+v<sup>3</sup>=-6</th><th>Z<sup>2</sup>+6Z+<math>(\frac{7}{3})^3</math>=0</th><th>Z=-3±<math>\frac{10i}{3\sqrt{3}}</math></th><th>u,v=1±<math>\frac{2}{\sqrt3}i</math></th><th>u,v=<math>\frac{2}{2}</math>±<math>\frac{2+2}{2\sqrt{3}}i</math></th></tr>

<tr><th>n=3 </th><th>1,3,-4</th><th>x<sup>3</sup>-13x+12=0</th><th>uv=13/3,u<sup>3</sup>+v<sup>3</sup>=-12</th><th>Z<sup>2</sup>+12Z+<math>(\frac{13}{3})^3</math>=0</th><th>Z=-6±<math>\frac{35i}{3\sqrt{3}}</math></th><th>u,v=<math>\frac{3}{2}</math>±<math>\frac{2.5}{\sqrt3}i</math></th><th>u,v=<math>\frac{3}{2}</math>±<math>\frac{3+2}{2\sqrt{3}}i</math></th></tr>

<tr><th>n=4 </th><th>1,4,-5</th><th>x<sup>3</sup>-21x+20=0</th><th>uv=7,u<sup>3</sup>+v<sup>3</sup>=-20</th><th>Z<sup>2</sup>+20Z+<math>7^3</math>=0</th><th>Z=-10±<math>9\sqrt{3}i</math></th><th>u,v=2±<math>\sqrt3i</math></th><th>u,v=<math>\frac{4}{2}</math>±<math>\frac{4+2}{2\sqrt{3}}i</math></th></tr>

</table>
==解法一：卡爾丹法==
===階段一：變形去除三次項===
1.<math>at^3 + bt^2 + ct + d = 0 \Rightarrow </math> 以 <math>t = x - \frac{b}{3a}</math> 代入

2.得 <math>x^3 - 3px + q = 0 </math>，令其三根為 <math>x_1,x_2,x_3</math>

===階段二：變身為二次方程式===
3.<table><tr><th>令<math>\begin{cases}
x_1=u + v \\
x_2=u \omega + v \omega^2 \\
x_3=u \omega^2 + v \omega
\end{cases}</math></th><th><math> \Rightarrow </math></th><th><math>(x-x_1)(x-x_2)(x-x_3)=0</math></th><th><math> \Rightarrow </math></th><th><math>\begin{cases}
x_1+x_2+x_3=0 \\
x_1x_2+x_1x_3+x_2x_3=-3uv=-3p \\
-x_1x_2x_3=-(u^3+v^3)=q
\end{cases}</math></th></tr></table>

4.<table><tr><th>設<math>\begin{cases}
Z_1=u^3 \\
Z_2=v^3
\end{cases}</math></th><th><math> \Rightarrow </math></th><th><math>\begin{cases}
Z_1+Z_2=-q \\
Z_1Z_2=(p)^3
\end{cases}</math></th></tr></table>

5.故 <math>Z_1,Z_2</math> 為 <math>Z^2+qZ+(p)^3=0</math> 的兩根

===階段三：以二次方程式之兩根求三次方程式之三根===
6.<table><tr><th><math>u=\sqrt[3]{Z_1} , v=\sqrt[3]{Z_2}</math></th><th><math>\begin{cases}
x_1=u + v \\
x_2=u \omega + v \omega^2 \\
x_3=u \omega^2 + v \omega
\end{cases}</math></th></tr></table>

==解法二：參數法==
===階段一：同卡爾單法的階段一===
得 x<sup>3</sup>-3px+q=0
#一元三次方程式必有一實根，改變單位，重新將此實根定義為 1 單位。
#另一根與此根的比值定為 n 則三根為 1,n,-(n+1)

===階段二：變身為二次方程式===
原式=(x-1)(x-n)(x+n+1)=X<sup>3</sup>-(n<sup>2</sup>+n+1)x+n(n+1)=0

q=n(n+1),3p=n(n+1)+1

<math>Z^2+qZ+(p)^3=0=Z^2+n(n+1)Z+[\frac{n(n+1)+1}{3}]^3</math>

<math>Z=\frac{-n(n+1)}{2}</math>±<math>\sqrt{[\frac{n(n+1)}{2}]^2-[\frac{n(n+1)+1}{3}]^3}</math>

令 <math>f(n)=[\frac{n(n+1)}{2}]^2-[\frac{n(n+1)+1}{3}]^3 \\

f(n)=\frac{-1}{108}(4n^6+12n^5-3n^4-26n^3-3n^2+12n+4) \\

f(n)=\frac{-1}{108}(n-1)^2(2n+1)^2(n+2)^2 \\

\sqrt{f(n)}=\frac{i}{6\sqrt3}(n-1)(2n+1)(n+2)</math>

所以

<math>Z=\frac{-n(n+1)}{2} \pm \frac{(n-1)(2n+1)(n+2)}{6\sqrt3}i \\

=\frac{1}{24\sqrt3}[-3\sqrt3\cdot4\cdot n(n+1)\pm(-8n^3-12n^2+12n+8)i] \\

=\frac{1}{24\sqrt3}[(3\sqrt3n^3-3\sqrt3n(n^2+4n+4))\pm(n^3+6n^2+12n+8-9n^3-18n^2)i] \\

=(\frac{1}{2\sqrt3})^3(\sqrt3n\pm(n+2)i)^3 =[\frac{1}{2\sqrt3}(\sqrt3n\pm(n+2)i)]^3=[\frac{n}{2}\pm\frac{n+2}{2\sqrt3}i]^3

u,v=\frac{n}{2}\pm\frac{n+2}{2\sqrt3}i</math>