Formula for the Roots of a Cubic Equation
📂Abstract Algebra Formula for the Roots of a Cubic Equation The solution of the cubic equation t 3 + p t + q = 0 t^{3}+pt+q = 0 t 3 + pt + q = 0
{ t 1 = u 1 + v 1 = − q 2 + q 2 4 + p 3 27 3 + − q 2 − q 2 4 + p 3 27 3 t 2 = u 2 + v 3 = − q 2 + q 2 4 + p 3 27 3 ω + − q 2 − q 2 4 + p 3 27 3 ω 2 t 3 = u 3 + v 2 = − q 2 + q 2 4 + p 3 27 3 ω 2 + − q 2 − q 2 4 + p 3 27 3 ω
\begin{cases} t_{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}}}
\\ t_{2}=u_{2}+v_{3}=\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}
\\ t_{3}=u_{3}+v_{2}=\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\end{cases}
⎩ ⎨ ⎧ t 1 = u 1 + v 1 = 3 − 2 q + 4 q 2 + 27 p 3 + 3 − 2 q − 4 q 2 + 27 p 3 t 2 = u 2 + v 3 = 3 − 2 q + 4 q 2 + 27 p 3 ω + 3 − 2 q − 4 q 2 + 27 p 3 ω 2 t 3 = u 3 + v 2 = 3 − 2 q + 4 q 2 + 27 p 3 ω 2 + 3 − 2 q − 4 q 2 + 27 p 3 ω
Here ω = e i 2 3 π \omega = e^{i\frac{2}{3}\pi} ω = e i 3 2 π
Proof Cardano’s Method Let us consider a cubic equation a x 3 + b x 2 + c x + d = 0 ( a ≠ 0 ) ax^{3}+bx^{2}+cx+d=0(a\ne0) a x 3 + b x 2 + c x + d = 0 ( a = 0 )
x 3 + a x 2 + b x + c = 0
\begin{equation}
x^{3}+ax^{2} +bx+c=0
\end{equation}
x 3 + a x 2 + b x + c = 0
To eliminate the quadratic term, substitute x = t − a 3 x=t-{\textstyle \frac{a}{3}} x = t − 3 a
( t − a 3 ) 3 + a ( t − a 3 ) 2 + b ( t − a 3 ) + c = 0 ⟹ ( t 3 − a t 2 + a 2 3 t − a 3 27 ) + ( a t 2 − 2 a 2 3 t + a 3 9 ) + ( b t − a b 3 ) + c = 0 ⟹ t 3 + ( a 2 3 − 2 a 2 3 + b ) t + ( − a 3 27 + a 3 9 − a b 3 + c ) = 0 ⟹ t 3 + ( b − a 2 3 ) t + ( 2 a 3 27 − a b 3 + c ) = 0
\begin{align*}
&&\left( t- \frac{a}{3}\right)^{3}+a\left( t-\frac{a}{3} \right)^{2}+b\left( t-\frac{a}{3} \right) + c &=0
\\ \implies && \left( t^{3}-\cancel{at^{2}}+\frac{a^{2}}{3}t-\frac{a^{3}}{27} \right)+\left(\cancel{at^{2}}-\frac{2a^{2}}{3}t + \frac{a^{3}}{9}\right) + \left( bt-\frac{ab}{3} \right) +c &= 0
\\ \implies && t^{3}+\left( \frac{a^{2}}{3} -\frac{2a^{2}}{3}+b \right)t +\left( -\frac{a^{3}}{27}+\frac{a^{3}}{9}-\frac{ab}{3}+c \right)&=0
\\ \implies && t^{3}+\left( b- \frac{a^{2}}{3} \right)t +\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)&=0
\end{align*}
⟹ ⟹ ⟹ ( t − 3 a ) 3 + a ( t − 3 a ) 2 + b ( t − 3 a ) + c ( t 3 − a t 2 + 3 a 2 t − 27 a 3 ) + ( a t 2 − 3 2 a 2 t + 9 a 3 ) + ( b t − 3 ab ) + c t 3 + ( 3 a 2 − 3 2 a 2 + b ) t + ( − 27 a 3 + 9 a 3 − 3 ab + c ) t 3 + ( b − 3 a 2 ) t + ( 27 2 a 3 − 3 ab + c ) = 0 = 0 = 0 = 0
To further simplify the equation, let’s set p = b − a 2 3 p=b-\frac{a^{2}}{3} p = b − 3 a 2 q = 2 a 3 27 − a b 3 + c q=\frac{2a^{3}}{27}-\frac{ab}{3}+c q = 27 2 a 3 − 3 ab + c
t 3 + p t + q = 0
\begin{equation}
t^{3}+pt+q = 0
\end{equation}
t 3 + pt + q = 0
If we substitute once more with t = u + v t=u+v t = u + v
( u + v ) 3 + p ( u + v ) + q = 0
\begin{equation}
(u+v)^{3}+p(u+v)+q=0
\end{equation}
( u + v ) 3 + p ( u + v ) + q = 0
Also, according to the multiplication formula , the following equation holds.
( u + v ) 3 = u 3 + 3 u 2 v + 3 u v 2 + v 2 = u 3 + v 3 + 3 u v ( u + v )
(u+v)^{3}=u^{3}+3u^{2}v+3uv^{2}+v^{2}=u^{3}+v^{3}+3uv(u+v)
( u + v ) 3 = u 3 + 3 u 2 v + 3 u v 2 + v 2 = u 3 + v 3 + 3 uv ( u + v )
Transferring the entire right-hand side to the left yields the equation below.
( u + v ) 3 − 3 u v ( u + v ) − ( u 3 + v 3 ) = 0
\begin{equation}
(u+v)^{3} - 3uv(u+v) - (u^{3}+v^{3})=0
\end{equation}
( u + v ) 3 − 3 uv ( u + v ) − ( u 3 + v 3 ) = 0
Comparing ( 3 ) (3) ( 3 ) ( 4 ) (4) ( 4 ) ( 3 ) (3) ( 3 ) u u u v v v p = − 3 u v p=-3uv p = − 3 uv q = − ( u 3 + v 3 ) q=-(u^{3}+v^{3}) q = − ( u 3 + v 3 ) u u u v v v
u 3 v 3 = − p 3 27 u 3 + v 3 = − q
\begin{align*}
u^{3}v^{3} &= -\frac{p^{3}}{27}
\\ u^{3}+v^{3} &= -q
\end{align*}
u 3 v 3 u 3 + v 3 = − 27 p 3 = − q
Considering Vieta’s formulas for quadratic equations, we can see that u 3 u^{3} u 3 v 3 v^{3} v 3
X 2 + q X − p 3 27 = 0
X^{2} +qX -\frac{p^{3}}{27}=0
X 2 + qX − 27 p 3 = 0
Then, the quadratic formula yields the following equation.
X 1 = u 3 = − q + q 2 + 4 p 3 27 2 = − q 2 + q 2 4 + p 3 27 X 2 = v 3 = − q − q 2 + 4 p 3 27 2 = − q 2 − q 2 4 + p 3 27
\begin{equation}
\begin{aligned}
X_{1}&=u^{3}=\frac{-q+\sqrt{q^{2}+\frac{4p^{3}}{27}}}{2}=-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}
\\ X_{2}&= v^{3}=\frac{-q-\sqrt{q^{2}+\frac{4p^{3}}{27}}}{2}=-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}
\end{aligned}
\end{equation}
X 1 X 2 = u 3 = 2 − q + q 2 + 27 4 p 3 = − 2 q + 4 q 2 + 27 p 3 = v 3 = 2 − q − q 2 + 27 4 p 3 = − 2 q − 4 q 2 + 27 p 3
At this point, for any real number α \alpha α z 3 = α z^{3}=\alpha z 3 = α
z 1 = α 3 and z 2 = α 3 ω and z 3 = α 3 ω 2
z_{1}=\sqrt[3]{\alpha} \quad \text{and} \quad z_{2}=\sqrt[3]{\alpha}\omega \quad \text{and} \quad z_{3}=\sqrt[3]{\alpha}\omega^{2}
z 1 = 3 α and z 2 = 3 α ω and z 3 = 3 α ω 2
Here ω \omega ω ω 3 = 1 \omega^{3}=1 ω 3 = 1 ω = e i 2 3 π = cos ( 2 3 π ) + i sin ( 2 3 π ) \omega=e^{i\frac{2}{3}\pi}=\cos (\frac{2}{3}\pi)+i\sin (\frac{2}{3}\pi) ω = e i 3 2 π = cos ( 3 2 π ) + i sin ( 3 2 π ) u , v u, v u , v ( 5 ) (5) ( 5 )
{ u 1 = − q 2 + q 2 4 + p 3 27 3 u 2 = − q 2 + q 2 4 + p 3 27 3 ω u 3 = − q 2 + q 2 4 + p 3 27 3 ω 2 and { v 1 = − q 2 − q 2 4 + p 3 27 3 v 2 = − q 2 − q 2 4 + p 3 27 3 ω v 3 = − q 2 − q 2 4 + p 3 27 3 ω 2
\begin{cases} u_{1}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}
\\ u_{2}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega
\\ u_{3}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2}\end{cases}\quad \text{and} \quad
\begin{cases} v_{1}=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}
\\ v_{2}=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega
\\ v_{3}=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2}\end{cases}
⎩ ⎨ ⎧ u 1 = 3 − 2 q + 4 q 2 + 27 p 3 u 2 = 3 − 2 q + 4 q 2 + 27 p 3 ω u 3 = 3 − 2 q + 4 q 2 + 27 p 3 ω 2 and ⎩ ⎨ ⎧ v 1 = 3 − 2 q − 4 q 2 + 27 p 3 v 2 = 3 − 2 q − 4 q 2 + 27 p 3 ω v 3 = 3 − 2 q − 4 q 2 + 27 p 3 ω 2
However, since p = − u v p=-uv p = − uv
{ t 1 = u 1 + v 1 = − q 2 + q 2 4 + p 3 27 3 + − q 2 − q 2 4 + p 3 27 3 t 2 = u 2 + v 3 = − q 2 + q 2 4 + p 3 27 3 ω + − q 2 − q 2 4 + p 3 27 3 ω 2 t 3 = u 3 + v 2 = − q 2 + q 2 4 + p 3 27 3 ω 2 + − q 2 − q 2 4 + p 3 27 3 ω
\begin{cases} t_{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}}}
\\ t_{2}=u_{2}+v_{3}=\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}
\\ t_{3}=u_{3}+v_{2}=\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\end{cases}
⎩ ⎨ ⎧ t 1 = u 1 + v 1 = 3 − 2 q + 4 q 2 + 27 p 3 + 3 − 2 q − 4 q 2 + 27 p 3 t 2 = u 2 + v 3 = 3 − 2 q + 4 q 2 + 27 p 3 ω + 3 − 2 q − 4 q 2 + 27 p 3 ω 2 t 3 = u 3 + v 2 = 3 − 2 q + 4 q 2 + 27 p 3 ω 2 + 3 − 2 q − 4 q 2 + 27 p 3 ω
Since all cubic equations can be represented without a quadratic term as ( 2 ) (2) ( 2 ) ( 1 ) (1) ( 1 )
{ x 1 = − 1 2 ( 2 a 3 27 − a b 3 + c ) + 1 4 ( 2 a 3 27 − a b 3 + c ) 2 + 1 27 ( b − a 2 3 ) 3 3 + − 1 2 ( 2 a 3 27 − a b 3 + c ) − 1 4 ( 2 a 3 27 − a b 3 + c ) 2 + 1 27 ( b − a 2 3 ) 3 3 − a 3 x 2 = − 1 2 ( 2 a 3 27 − a b 3 + c ) + 1 4 ( 2 a 3 27 − a b 3 + c ) 2 + 1 27 ( b − a 2 3 ) 3 3 ω + − 1 2 ( 2 a 3 27 − a b 3 + c ) − 1 4 ( 2 a 3 27 − a b 3 + c ) 2 + 1 27 ( b − a 2 3 ) 3 3 ω 2 − a 3 x 3 = − 1 2 ( 2 a 3 27 − a b 3 + c ) + 1 4 ( 2 a 3 27 − a b 3 + c ) 2 + 1 27 ( b − a 2 3 ) 3 3 ω 2 + − 1 2 ( 2 a 3 27 − a b 3 + c ) − 1 4 ( 2 a 3 27 − a b 3 + c ) 2 + 1 27 ( b − a 2 3 ) 3 3 ω − a 3
\begin{cases} x_{1}=\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)+\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}+\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)-\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}-\frac{a}{3}
\\ x_{2}=\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)+\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega+\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)-\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega^{2} -\frac{a }{3}
\\ x_{3}=\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)+\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega^{2}+\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)-\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega -\frac{a }{3} \end{cases}
⎩ ⎨ ⎧ x 1 = 3 − 2 1 ( 27 2 a 3 − 3 ab + c ) + 4 1 ( 27 2 a 3 − 3 ab + c ) 2 + 27 1 ( b − 3 a 2 ) 3 + 3 − 2 1 ( 27 2 a 3 − 3 ab + c ) − 4 1 ( 27 2 a 3 − 3 ab + c ) 2 + 27 1 ( b − 3 a 2 ) 3 − 3 a x 2 = 3 − 2 1 ( 27 2 a 3 − 3 ab + c ) + 4 1 ( 27 2 a 3 − 3 ab + c ) 2 + 27 1 ( b − 3 a 2 ) 3 ω + 3 − 2 1 ( 27 2 a 3 − 3 ab + c ) − 4 1 ( 27 2 a 3 − 3 ab + c ) 2 + 27 1 ( b − 3 a 2 ) 3 ω 2 − 3 a x 3 = 3 − 2 1 ( 27 2 a 3 − 3 ab + c ) + 4 1 ( 27 2 a 3 − 3 ab + c ) 2 + 27 1 ( b − 3 a 2 ) 3 ω 2 + 3 − 2 1 ( 27 2 a 3 − 3 ab + c ) − 4 1 ( 27 2 a 3 − 3 ab + c ) 2 + 27 1 ( b − 3 a 2 ) 3 ω − 3 a
■
