背景

---

说到方程求解，我想大家都会想到一元二次方程：

<script type="math/tex; mode=display" id="MathJax-Element-148">ax^2 + bx + c = 0</script>
可使用 配方法，求得根为：
<script type="math/tex; mode=display" id="MathJax-Element-149">x = \frac {-b \pm \sqrt {b^2-4ac}} {2a}</script>

当 b2−4ac>0 <script type="math/tex" id="MathJax-Element-150">b^2 - 4ac > 0 </script>时，方程有两实根
当 b2−4ac=0 <script type="math/tex" id="MathJax-Element-151">b^2 - 4ac = 0</script>时，方程有两相同实根
当 b2−4ac<0 <script type="math/tex" id="MathJax-Element-152">b^2 - 4ac < 0</script>时，没有实根，但在复数上有两复根

当二次方程求解成功之后，人们开始转向研究解决三次主程的方法。然而在解决三次方程时，人们的进展没有想像中那么快。但人们还是找到了一些特殊三次方程的解决，于是乎，学术界和民间纷纷开展了一场解决三次方程的比赛。也有人找到了通用的解方程办法，但是秘而不宣，造成三次方程的解决一直没有告知于众。

数学界解三次方程这段历史非常有历，在这些就不详细说明了，有兴趣的读者可以网上google一下，可以找到不错的史料。

为了降低阅读的难度，本文先介绍一种特殊三次方程的解决，看看如何求解决这种特殊的三次方程。

特殊三次方程求解

---

卡尔丹首先突破的三次方程是没有二次项的三次方程，它的形式如下：

<script type="math/tex; mode=display" id="MathJax-Element-6">x^3 + px + q = 0</script>

那如何解这个新方程呢？卡尔丹想到很奇妙的方法，令 x=u+v <script type="math/tex" id="MathJax-Element-7">x=u+v</script>，代入原方程，化简后得：

<script type="math/tex; mode=display" id="MathJax-Element-8">u^3 + v^3 + 3uv(u+v) + p(u+v)+q = 0</script>

然后，再对u和v增加一个约束： 3uv=−p <script type="math/tex" id="MathJax-Element-9">3uv=-p</script>，就可以将 (u+v) <script type="math/tex" id="MathJax-Element-10">(u+v)</script>项消灭，从而得到：

<script type="math/tex; mode=display" id="MathJax-Element-11">u^3 + v^3 + q = 0</script>

由增加的约束可得 v=−p3u <script type="math/tex" id="MathJax-Element-12">v=-\frac {p} {3u}</script>，代入得：

<script type="math/tex; mode=display" id="MathJax-Element-13">u^3 + (-p/3u)^3 + q = 0</script>

然后两边同时乘以 27u3 <script type="math/tex" id="MathJax-Element-14">27u^3</script>，整理移项得：

<script type="math/tex; mode=display" id="MathJax-Element-15">27u^6 + 27qu^3 - p^3 = 0</script>

显然，此方程可以看作是 u3 <script type="math/tex" id="MathJax-Element-16">u^3</script>为变元的二次方程，根据二次方程的求根公式，马上可以得到：

<script type="math/tex; mode=display" id="MathJax-Element-17">u^3 = -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}</script>

从这里开始就犯难了，这里遇到两个问题：
1. 如果 q24+p327<0 <script type="math/tex" id="MathJax-Element-18">\frac {q^2} {4} + \frac {p^3} {27} < 0</script>，则 q24+p327−−−−−−√ <script type="math/tex" id="MathJax-Element-19">\sqrt {\frac {q^2} {4} + \frac {p^3} {27}}</script>是复数。
2. 从 u3 <script type="math/tex" id="MathJax-Element-20">u^3</script>中求解 u <script type="math/tex" id="MathJax-Element-21">u</script>，似乎又涉及3次方程求解，它有3个根，而且涉及复数。

为了保证求解过程不产生跳跃性，我们对q24+p327<script type="math/tex" id="MathJax-Element-22">\frac {q^2} {4} + \frac {p^3} {27} </script>的值分情况讨论。

情况一： q24+p327≥0 <script type="math/tex" id="MathJax-Element-23">\frac {q^2} {4} + \frac {p^3} {27} \ge 0</script>

因为从 u3 <script type="math/tex" id="MathJax-Element-24">u^3</script>求解 u <script type="math/tex" id="MathJax-Element-25">u</script>，涉及开3次方，而3次方根却一定涉及到复数，对于没有学习过复变函数的读者来说，会有一定的困难。

即然涉及开3次方，那么我们将u3<script type="math/tex" id="MathJax-Element-26">u^3</script>写到复数形式，同时为了方便做开方运算，使用复数的三角形式。

<script type="math/tex; mode=display" id="MathJax-Element-27">u^3 = -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}} + 0i =( -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}} )(cos 0 + isin 0)</script>

根据复数的定义，开3次方，等于模开3次方，而辐解有3个值，即：

<script type="math/tex; mode=display" id="MathJax-Element-28">u = (\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}})(cos \frac {0 + 2n\pi} {3} + isin \frac {0 + 2n\pi} 3) </script>
其中 n=0,1,3 <script type="math/tex" id="MathJax-Element-29">n = 0, 1, 3</script>

即

<script type="math/tex; mode=display" id="MathJax-Element-30">u_1 = \sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}</script>
<script type="math/tex; mode=display" id="MathJax-Element-31">u_2 = (\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega</script>
<script type="math/tex; mode=display" id="MathJax-Element-32">u_3 = (\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2</script>

其中 ω=cos2π3+isin2π3=−12+3√2i <script type="math/tex" id="MathJax-Element-33">\omega = cos \frac {2\pi} 3 + i sin \frac {2\pi} 3 = -\frac 1 2 + {\frac {\sqrt 3} 2}i</script>

接下来，就可以计算 v <script type="math/tex" id="MathJax-Element-34">v</script>了。 由3uv=−p<script type="math/tex" id="MathJax-Element-35">3uv = -p</script> 得到:

<script type="math/tex; mode=display" id="MathJax-Element-36">v = -\frac {p} {3u}</script>

取 u1=−q2±q24+p327−−−−−−√−−−−−−−−−−−−−√3 <script type="math/tex" id="MathJax-Element-37">u_1 = \sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}</script>

可得

<script type="math/tex; mode=display" id="MathJax-Element-38">v_1 = -\frac p {3\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}} </script> <script type="math/tex; mode=display" id="MathJax-Element-39">= -\frac {p \sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}} {3(\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}})(\sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}})} </script> <script type="math/tex; mode=display" id="MathJax-Element-40">= -\frac {p \sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}}{3\sqrt[3] {\frac {q^2} 4 - (\frac {q^2} 4 + \frac {p^3} {27})}} </script> <script type="math/tex; mode=display" id="MathJax-Element-41">=-\frac {p\sqrt[3] {-\frac q 2 \mp \sqrt {\frac {q^2} 4 + \frac {p^3} {27}}}} {3\sqrt[3] {-\frac {p^3} {27}}} </script> <script type="math/tex; mode=display" id="MathJax-Element-42">= \sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}</script>

有没有从 u1 <script type="math/tex" id="MathJax-Element-43">u_1</script>求 v1 <script type="math/tex" id="MathJax-Element-44">v_1</script>计算过程中，发现这个计算过程很有规律，没有丝毫的复杂性，一切看起来很简洁。

由 u2=(−q2±q24+p327−−−−−−√−−−−−−−−−−−−−√3)ω <script type="math/tex" id="MathJax-Element-45">u_2 = (\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega</script>，可得：

<script type="math/tex; mode=display" id="MathJax-Element-46">v_2 = -\frac p {3 (\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega} </script> <script type="math/tex; mode=display" id="MathJax-Element-47">= -\frac p {3 \sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}} } (\frac 1 \omega) </script> <script type="math/tex; mode=display" id="MathJax-Element-48">= (\sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) (\frac 1 \omega) </script> <script type="math/tex; mode=display" id="MathJax-Element-49">=(\sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2</script>

同理，由 u3=(−q2±q24+p327−−−−−−√−−−−−−−−−−−−−√3)ω2 <script type="math/tex" id="MathJax-Element-50">u_3 = (\sqrt[3] { -\frac q 2 \pm \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2</script>
可得

<script type="math/tex; mode=display" id="MathJax-Element-51">v_3 = (\sqrt[3] { -\frac q 2 \mp \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega</script>

由前面引入 u <script type="math/tex" id="MathJax-Element-52">u</script>和v<script type="math/tex" id="MathJax-Element-53">v</script>时，有 x=u+v <script type="math/tex" id="MathJax-Element-54">x = u + v</script>关系，而在求解过程中发现 u <script type="math/tex" id="MathJax-Element-55">u</script>和v<script type="math/tex" id="MathJax-Element-56">v</script>分别有6组值，是不是说 x <script type="math/tex" id="MathJax-Element-57">x</script>也有6组值呢？非也，如何细仔观察这6组值，会发现u<script type="math/tex" id="MathJax-Element-58">u</script>和 v <script type="math/tex" id="MathJax-Element-59">v</script>有对称性，他们加起来一才有3组值，分别是：

<script type="math/tex; mode=display" id="MathJax-Element-60">x_1 = u_1 + v_1 = \sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}+\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}</script>

<script type="math/tex; mode=display" id="MathJax-Element-61">x_2 = u_2 + v_2 = (\sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega + (\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2</script>
<script type="math/tex; mode=display" id="MathJax-Element-62">x_3 = u_3 + v_3 = (\sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2 + (\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega</script>

注意：这里的 √ <script type="math/tex" id="MathJax-Element-63">\sqrt {}</script>是求算术平方根，运行结果只取正值，而 √3 <script type="math/tex" id="MathJax-Element-64">\sqrt[3] {}</script>是在实数范围内求3次方根，运行结果为实数结果。

情况二： q24+p327<0 <script type="math/tex" id="MathJax-Element-65">\frac{q^2} {4} + \frac {p^3} {27} \lt 0 </script>

如果 q24+p327<0 <script type="math/tex" id="MathJax-Element-66">\frac{q^2} {4} + \frac {p^3} {27} < 0 </script>，则开平方根是一个纯虚数，那应该写成：

<script type="math/tex; mode=display" id="MathJax-Element-67">u^3 = -\frac q 2 \pm (\sqrt {-\frac{q^2} {4} - \frac {p^3} {27}})i</script>

注意：这个式中的 √ <script type="math/tex" id="MathJax-Element-68">\sqrt {}</script>仍然是算术平方根。

为了方便开3次方求解 u <script type="math/tex" id="MathJax-Element-69">u</script>，我们将u3<script type="math/tex" id="MathJax-Element-70">u^3</script>写成它的三角形式，要写它的三角形式。由于 u3 <script type="math/tex" id="MathJax-Element-71">u^3</script>实部和虚部同时存在，则需要求 u3 <script type="math/tex" id="MathJax-Element-72">u^3</script>的模和辐角。

<script type="math/tex; mode=display" id="MathJax-Element-73">|u^3| = \sqrt {(-\frac q 2)^2 + (\sqrt {-\frac {q^2} 4 - \frac {p^3} {27}})^2 } = \sqrt {-\frac {p^3} {27}}</script>

接下来要求辐角了，为了避免后续再从三角形式转换成代数形式，这里不对辐解进行求解，直接将它的辐角记为 θ <script type="math/tex" id="MathJax-Element-74">\theta</script> ，满足：

<script type="math/tex; mode=display" id="MathJax-Element-75">-\frac \pi 2 \le \theta \le \frac \pi 2 </script>
<script type="math/tex; mode=display" id="MathJax-Element-76">cos\theta = \frac {-\frac q 2} {\sqrt {-\frac {p^3} {27}}}</script>
<script type="math/tex; mode=display" id="MathJax-Element-77">sin\theta = \frac {\sqrt {-\frac{q^2} {4} - \frac {p^3} {27}}} {\sqrt {-\frac {p^3} {27}}}</script>

那么 u3 <script type="math/tex" id="MathJax-Element-78">u^3</script>可表示为：

<script type="math/tex; mode=display" id="MathJax-Element-79">u^3 = \sqrt {-\frac {p^3} {27}}(cos\theta + isin\theta)</script>
从而有：
<script type="math/tex; mode=display" id="MathJax-Element-80">u = \sqrt[3] {\sqrt {-\frac {p^3} {27}}}(cos \frac {\theta + 2n\pi} {3} + isin \frac {\theta + 2n\pi} {3})=\sqrt {-\frac {p} {3}} (cos \frac {\theta + 2n\pi} {3} + isin \frac {\theta + 2n\pi} {3}) (n = 0, 1,2)</script>

根据复数的定义： (cosα+isinα)(cosβ+isinβ)=cos(α+β)+isin(α+β) <script type="math/tex" id="MathJax-Element-81">(cos\alpha + isin\alpha)(cos\beta + isin\beta) = cos (\alpha + \beta) + isin(\alpha + \beta)</script>，将 u <script type="math/tex" id="MathJax-Element-82">u</script>表达式后面的复数部分拆成两个复数相乘，即：

<script type="math/tex; mode=display" id="MathJax-Element-83">u=\sqrt {- \frac p 3} (cos \frac \theta 3 + isin \frac \theta 3) (cos \frac {2n\pi} {3} + isin \frac {2n\pi} {3}) = \sqrt {- \frac p 3} (cos \frac \theta 3 + isin \frac \theta 3)\omega^n (n = 0, 1,2)</script>

怎么样， ω <script type="math/tex" id="MathJax-Element-84">\omega</script>又现身了，求三次方根，永远都少不了 ω <script type="math/tex" id="MathJax-Element-85">\omega</script>的身影，正如对负数求平方根一样，永远少不了 i <script type="math/tex" id="MathJax-Element-86">i</script>的身影一样。

为了减少对θ<script type="math/tex" id="MathJax-Element-87">\theta</script>的依赖，我们先消去 θ <script type="math/tex" id="MathJax-Element-88">\theta</script>再计算 v <script type="math/tex" id="MathJax-Element-89">v</script>。由于有θ<script type="math/tex" id="MathJax-Element-90">\theta</script>，如何将它变成 θ <script type="math/tex" id="MathJax-Element-91">\theta</script>呢，很简单，先做立方运算，再开3次方。即：

<script type="math/tex; mode=display" id="MathJax-Element-92">u = \sqrt {- \frac p 3} (cos \frac \theta 3 + isin \frac \theta 3)\omega^n </script> <script type="math/tex; mode=display" id="MathJax-Element-93">= \sqrt[3] {(\sqrt {- \frac p 3})^3(cos \frac \theta 3 + isin \frac \theta 3)^3} \omega^n </script> <script type="math/tex; mode=display" id="MathJax-Element-94">= \sqrt[3] {\sqrt {- \frac {p^3} {27}}(cos \theta + isin \theta }) \omega^n </script> <script type="math/tex; mode=display" id="MathJax-Element-95">=\sqrt[3] { -\frac q 2 + \sqrt {-\frac {q^2} {4} - \frac {p^3} {27}}}i \omega^n</script>

噢，上式中的 −q2+−q24−p327−−−−−−−−√−−−−−−−−−−−−−−−√3i <script type="math/tex" id="MathJax-Element-96">\sqrt[3] {-\frac q 2 + \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}}i</script>又变回了对复数开三次方，但了复数原理的都知道，任一复数开三次方根时，有三个数值，那么这里的具体表示哪一个呢？

答案是任一个都可以，但这个根值选定之后，后面所有 −q2+−q24−p327−−−−−−−−√−−−−−−−−−−−−−−−√3i <script type="math/tex" id="MathJax-Element-97">\sqrt[3] {-\frac q 2 + \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}}i</script>结果都要使用相同的值，才能保证结果是正确的。

好了，下面由 3uv=−p <script type="math/tex" id="MathJax-Element-98">3uv = -p</script>，从 u <script type="math/tex" id="MathJax-Element-99">u</script>求解v<script type="math/tex" id="MathJax-Element-100">v</script>：

<script type="math/tex; mode=display" id="MathJax-Element-101">v = -\frac {p} {3u} </script> <script type="math/tex; mode=display" id="MathJax-Element-102">= - \frac {p} {3\sqrt[3] { -\frac q 2 + \sqrt {-\frac {q^2} {4} - \frac {p^3} {27}}}i \omega^n} </script> <script type="math/tex; mode=display" id="MathJax-Element-103">= - \frac {p(\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i})} {3(\sqrt[3] { -\frac q 2 + \sqrt {-\frac {q^2} {4} - \frac {p^3} {27}}i})(\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i})\omega^n} </script> <script type="math/tex; mode=display" id="MathJax-Element-104">=-\frac {p(\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i})} {3\sqrt[3] {\frac {q^2} 4 + (-\frac {q^2} 4 - \frac {p^3} {27})} \omega^n} </script> <script type="math/tex; mode=display" id="MathJax-Element-105">= - \frac {p(\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i})} {3\sqrt[3] {-\frac {p^3} {27}}\omega^n} </script> <script type="math/tex; mode=display" id="MathJax-Element-106"> = \frac {\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i}} {\omega^n} </script> <script type="math/tex; mode=display" id="MathJax-Element-107"> =(\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i})\omega^{3-n} (n = 0, 1,2) </script>

然后通过 x=u+v <script type="math/tex" id="MathJax-Element-108">x = u +v</script>，对 x <script type="math/tex" id="MathJax-Element-109">x</script>做运行有：

<script type="math/tex; mode=display" id="MathJax-Element-110">x = u + v = \sqrt[3] { -\frac q 2 + \sqrt {-\frac {q^2} {4} - \frac {p^3} {27}}}i \omega^n + (\sqrt[3] {-\frac q 2 - \sqrt {-\frac {q^2} 4 - \frac {p^3} {27}}i})\omega^{3-n} (n = 0, 1, 2) </script>

为了方便公式统一化，我们将算术平方根 √ <script type="math/tex" id="MathJax-Element-111">\sqrt {} </script>符号在复数上做一个扩展，即当n为负数，那么 n√ <script type="math/tex" id="MathJax-Element-112">\sqrt {n}</script>表示 −n−−−√i <script type="math/tex" id="MathJax-Element-113">\sqrt {-n}i</script>，如果n为正数是， n√ <script type="math/tex" id="MathJax-Element-114">\sqrt {n}</script>仍时算述平方根。同样地，对于开3次方根，会有3个根值，但我们使用 q√3 <script type="math/tex" id="MathJax-Element-115">\sqrt[3] q</script>表示为辐角值为 −π2≤θ≤π2 <script type="math/tex" id="MathJax-Element-116">-\frac \pi 2 \le \theta \le \frac \pi 2</script>的那个根。

那么求根分式可以写成：

<script type="math/tex; mode=display" id="MathJax-Element-117">x = (\sqrt[3] { -\frac q 2 + \sqrt {\frac {q^2} {4} + \frac {p^3} {27}}}) \omega^n + (\sqrt[3] {-\frac q 2 - \sqrt {\frac {q^2} 4 + \frac {p^3} {27}}})\omega^{3-n} (n = 0, 1, 2)</script>

综合两种情况下的求根据公式

---

上述两种情况下的求根公式其它是一样的，只是采用不同的写法而已。如果采用3根的表达方式时，写成如下：

<script type="math/tex; mode=display" id="MathJax-Element-153">x_1 = \sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}+\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}</script>
<script type="math/tex; mode=display" id="MathJax-Element-154">x_2 = (\sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega + (\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2</script>
<script type="math/tex; mode=display" id="MathJax-Element-155">x_3 = (\sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2 + (\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega</script>

其实还有情况二写成的那种通用公式，如下：

<script type="math/tex; mode=display" id="MathJax-Element-156">x = (\sqrt[3] { -\frac q 2 + \sqrt {\frac {q^2} {4} + \frac {p^3} {27}}} )\omega^n + (\sqrt[3] {-\frac q 2 - \sqrt {\frac {q^2} 4 + \frac {p^3} {27}}})\omega^{3-n} (n = 0, 1, 2)</script>

这里的秘密是 ω3=1 <script type="math/tex" id="MathJax-Element-157">\omega^3 = 1</script>，上述两组公式是等价的。但这个求解公式有些要注意的地方，这也是我第一次看到这个求根公式百思不得其解的原因：

1. 算术平方根运算，它里面是负数时，它的结果是前面带正号的纯虚数 （事实上用前带负号的纯虚数也可以，但后面的同值计算要保持一致）
2. 对于公式中的两个立方根运算，可以取3个结果中的一个，但一个确定后，另一个也随之确定，两者的结果需要满足 (−q2+q24+p327−−−−−−√−−−−−−−−−−−−−√3)(−q2−q24+p327−−−−−−√−−−−−−−−−−−−−√3)=−p3 <script type="math/tex" id="MathJax-Element-158"> (\sqrt[3] { -\frac q 2 + \sqrt {\frac {q^2} {4} + \frac {p^3} {27}}} )(\sqrt[3] {-\frac q 2 - \sqrt {\frac {q^2} 4 + \frac {p^3} {27}}}) = -\frac p 3</script>，为了方便计算，我们通常使用三次方根运行来表示辐角满足 −π2≤θ≤π2 <script type="math/tex" id="MathJax-Element-159">-\frac \pi 2 \le \theta \le \frac \pi 2</script>r 那个根

求根公式

---

综合上述两种情况，然后定义好二次方根和三次方根运行，求根公式可以表示为：

<script type="math/tex; mode=display" id="MathJax-Element-125">x_1 = \sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}+\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}</script>
<script type="math/tex; mode=display" id="MathJax-Element-126">x_2 = (\sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega + (\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2</script>
<script type="math/tex; mode=display" id="MathJax-Element-127">x_3 = (\sqrt[3] { -\frac q 2 + \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega^2 + (\sqrt[3] { -\frac q 2 - \sqrt {\frac{q^2} {4} + \frac {p^3} {27}}}) \omega</script>