Solution for a depressed cubic equation
Girolamo Cardano was an Italian mathematician who is famous for his book Ars Magna (The Great Art) on algebra. In his book, Cardano introduced the method of algebraic solution for the cubic and quartic equations which he had learnt from Tartaglia.
Cardano's Method is a standard way to solve cubic equations, which starts with cubics in the form of
The cubic equation without quadratic term is called depressed cubic equation.
Classic method to solve the depressed cubic equation
The basic idea behind the solution is to assume that x is a sum or difference of two variables so that the equation could be transformed to find the solution of the two variables. Hence, the value of x is determined. Once the depressed cubic equation is solved, more generic form of equation could be solved by algebraic manipulation.
Let
Take cubic of both sides,
x^3=(u-t)^3
=u^3-3u^2t+3ut^2-t^3
=-3ut(u-t)+u^3-t^3
=-3utx+u^3-t^3
Rearranging the terms, we get an identical depressed cubic equation like (1)
Comparing the equations (1) and (3), we get
Substituting (4) into (5) yields the equation.
This equation looks like a quadratic equation in terms of u^3 , which allows us to determine u through solving the quadratic equation and taking the cubic root subsequently.
Let m = u^3, equation (6) is simplified to
q =\dfrac{p^3}{27m}-m
Solving the quadratic equation yields
Plug in the result into equation (5) to obtain,
t^3=q+u^3=q+\bigg( -\dfrac{q}{2}\pm\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } \bigg)
=\dfrac{q}{2} \pm\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }
Therefore,
x=u-t = \sqrt[3]{-\dfrac{q}{2}\pm\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } -\sqrt[3]{\dfrac{q}{2} \pm\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}
Although it looks x has two solutions,
x_1= \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } -\sqrt[3]{\dfrac{q}{2} +\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}
and
x_2= \sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } -\sqrt[3]{\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}
they are actually equal to each other. Since -1 = \sqrt[3]{-1} for cubic root, we can move the minus sign of radical terms of both x_1 and x_2 into the radicals, which results in two equal results.
The formula looks complicated. But the radicants under the cubic root radicals of both terms are in good symmetry. If we denote the radicant of the first term as R, that is
R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }
the second radicant is then denoted as \overline{R} by taking the concept of conjugate.
R\cdotp\overline{R} is the perfect form of difference of squares.
R\cdotp\overline{R} = \bigg( -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }\bigg) \bigg( -\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }\bigg)= -\dfrac{p^3}{27}
And the reciprocal of R also correlates to the value of \overline{R}
\dfrac{1}{R}=\dfrac{1}{R} \cdotp \dfrac{\overline{R}}{\overline{R}} =-\dfrac{27}{p^3}\cdotp \overline{R}
Comparing the solution (9) with the assumption of (2), we can get
u =\sqrt[3]{R} , t = -\sqrt[3]{\overline{R}}
The denotation will frequently appear and be used in the later of the article to simplify the expression. At this point, we have concluded the solution of real numbers for a depressed cubic equation.
## Find the cube roots of 1
However, we are told there are 3 solutions for cubic equations, which possibly include complex numbers. Cardano didn't give what the three solutions look like since he didn't understand imaginary numbers at that time. Before we start to solve the equation with the cubics using the complex number. First, let's find the cube roots of 1. In order to understand the cubic roots in the field of complex numbers. Let's solve the equation
x^3=1
The solution for this simple equation is apparent if considering the solution of only real numbers. However, there are more to reveal when we stand in the field of complex numbers.
Move 1 to the same side of x^3,
x^3-1=0
Apply the difference of cubic formula,
(x-1)(x^2+x+1)=0
Then, we get one real solution and one quadratic equation
x- 1=0 \to x=1 or x^2+x+1=0
Solving the quadratic equation gives
x=\dfrac{-1\pm\sqrt{1-4} }{2} =\dfrac{-1\pm i\sqrt{3} }{2}
It's found there are 3 cube roots of 1 for the simple cubic equation, one real number and two conjugate imaginary numbers, namely, 1, \dfrac{-1+i\sqrt{3} }{2} and \dfrac{-1-i\sqrt{3}}{2}
Properties of the conjugate complex numbers
Besides the length or the modulus of 1, the two conjugate complex numbers exhibit interesting propertie. Let’s denote
ω = \dfrac{-1+i\sqrt{3}}{2} and \overline{ω} =\dfrac{-1-i\sqrt{3}}{2}
Taking square of ω gives
ω^2 =(\dfrac{-1+i\sqrt{3}}{2})^2
=\dfrac{1}{4}-\dfrac{i\sqrt{3} }{2} -\dfrac{3}{4}
=-\dfrac{1}{2}-\dfrac{i\sqrt{3} }{2}= \overline{ω}
which tells us the square of the complex number \dfrac{-1+i\sqrt{3}}{2} is equal to its conjugate.
Let’s have another manipulation by taking reciprocal of ω
\dfrac{1}{ω} =\dfrac{1}{\dfrac{-1+i\sqrt{3}}{2}}
=\dfrac{2}{-1+i\sqrt{3} } \cdotp \dfrac{-1-i\sqrt{3} }{-1-i\sqrt{3} }
=\dfrac{2}{1+3} \cdotp (-1-i\sqrt{3} )
=-\dfrac{-1-i\sqrt{3} }{2} =\overline{ω}
Similarly, it’s found that the reciprocal of the complex number \dfrac{-1+i\sqrt{3}}{2} is also equal to its conjugate.
Find the cube roots of x^3=A.
Furthermore, let's solve the cubic equation with only constant term.
Moving a to the same side as the variable
x^3-A=0
x^3-(\sqrt[3]{A} )^3=0
(x-\sqrt[3]{A} )(x^2 +\sqrt[3]{A}x+(\sqrt[3]{A})^2 )=0
Then we get
x_{1,2} = \dfrac{\sqrt[3]{A}\pm\sqrt{(\sqrt[3]{A})^2-4(\sqrt[3]{A})^2} }{2}
= \dfrac{-1\pm\sqrt{-3} }{2}\cdotp \sqrt[3]{A}
which shows that the second and the three solutions are the product of the conjugate complex numbers and a real root.
Solving the cubic using Vieta's Substitution
Let's come back to our initial steps of derivation. Substitute (4) into (2) gives
Substituting of the expression into (1) directly is called Vieta's Substitution, which effectively transforms the depressed cubic equation to a quadratic equation in terms of u^3. The substitution equation (10) shows that the second term is the product of the reciprocal of the first term u and -\dfrac{p}{3}
\Big( u-\dfrac{p}{3u}\Big) ^3+p\Big( u-\dfrac{p}{3u}\Big) +q=0
u^3-up+\dfrac{p^2}{3u}-\dfrac{p^3}{27u^3}+up-\dfrac{p^2}{3u}+q=0
Simplification yields the same equation as (6)
u^3-\dfrac{p^3}{27u^3}+q=0
Solving the equation gives the same solution for u^3 as (8)
Solving this cubic equation is the same as solving x^3=A
Using the results of (11) and (12), the cubic roots of u^3 are given directly as follow,
u_0 = \sqrt[3]{A}
u_1 = \dfrac{-1+ i\sqrt{3} }{2}\cdotp \sqrt[3]{A}
u_2 = \dfrac{-1- i\sqrt{3} }{2}\cdotp \sqrt[3]{A}
in which,
In Vieta's substitution equation (13), in order to obtain the solution of x, there left the second term -\dfrac{p}{3u} yet to be determined. Since A has two values (minus or plus before square root), there are total 6 cases to be evaluated plus 3 values of u. If using the denotations defined after (9), the value of A is either R or \overline{R}. The properties between R and \overline{R} are frequently used in the following derivation.
1) If A = R and u = \sqrt[3]{A}
-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp \dfrac{1}{ \sqrt[3]{R}}
= -\dfrac{p}{3}\cdotp \dfrac{1}{ \sqrt[3]{R}} \cdotp \dfrac{ \sqrt[3]{\overline{R}}}{ \sqrt[3]{\overline{R}} }
= -\dfrac{p}{3}\cdotp \dfrac{ \sqrt[3]{\overline{R}}}{\sqrt[3]{-\dfrac{p3}{27} } } =\sqrt[3]{\overline{R}}
Using Vieta's substitution equation (13)
x_1=u-\dfrac{p}{3u}= \sqrt[3]{R}+\sqrt[3]{\overline{R}}= \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } }
The result is exactly the same as that of (9).
2) If A = \overline{R} and u = \sqrt[3]{A}
-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp \dfrac{1}{\sqrt[3]{ \overline{R} } }
= -\dfrac{p}{3}\cdotp \dfrac{1}{\sqrt[3]{ \overline{R} } } \cdotp \dfrac{\sqrt[3]{R} }{\sqrt[3]{R} } =\sqrt[3]{R}
Substituting to Vieta's substitution equation yields the same result as the 1st case.
x_2=u-\dfrac{p}{3u} =\sqrt[3]{\overline{R} }+\sqrt[3]{R }
3) If A = R and u = \dfrac{-1+ i\sqrt{3} }{2}\cdotp \sqrt[3]{A}
-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp\dfrac{1}{ ω\cdotp \sqrt[3]{R}} = \overline{ω}\cdotp \sqrt[3]{\overline{R} }
x_3=u-\dfrac{p}{3u} = ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} }
4) If A = \overline{R} and u = \dfrac{-1+ i\sqrt{3} }{2}\cdotp \sqrt[3]{A}
-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp\dfrac{1}{ω\cdotp \sqrt[3]{\overline{R}}} = \overline{ω}\cdotp \sqrt[3]{R}
x_4=u-\dfrac{p}{3u} =ω\cdotp\sqrt[3]{\overline{R} }+ \overline{ω}\cdotp \sqrt[3]{R }
5) If A = R and u = \dfrac{-1- i\sqrt{3} }{2}\cdotp \sqrt[3]{A}
-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp\dfrac{1}{\overline{ω}\cdotp \sqrt[3]{R}} = ω\cdotp\sqrt[3]{\overline{R} }
x_5=u-\dfrac{p}{3u} = \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R} }
6) If A = \overline{R} and u = \dfrac{-1- i\sqrt{3} }{2}\cdotp \sqrt[3]{A}
-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp\dfrac{1}{ \overline{ω}\cdotp \sqrt[3]{\overline{R}}} =ω\cdotp \sqrt[3]{R }
x_6=u-\dfrac{p}{3u} = \overline{ω}\cdotp \sqrt[3]{\overline{R} }+ ω\cdotp \sqrt[3]{R }
It's found 3 pairs of cases result in the same results. In summary, there are three solutions for the depressed cubic equation, which are
Solving the cubic by factorization
Recall how we reach the cubic equation (3)
which is based on the condition x=u-t. In other words, u-t is one of solutions of equation (3). [x-(u-t)] is one of factors if we have factored the equation to the product of polynomials. Assuming the cubic equation (3) is able to be factored to the following form considering the known factor.
in which k and c is what we will need to solve out, representing the coefficient of linear and constant terms of the second factor.
Distribute the items of (16)
x^3+kx^2+cx-(u-t)x^2 -k(u-t)x-c(u-t) = 0
Combine the like terms
Comparing the terms of (17) with (3) gives,
and
Substituting (18) and (19) to (16) makes it clear of the second factor. Hence, the following equation is obtained,
x^2+(u-t)x+u^2+ut+t^2=0
Using the formula for the solution of general quadratic equation,
x = \dfrac{-(u-t)\pm\sqrt{(u-t)^2-4(u^2+ut+t^2)} }{2}
= \dfrac{-(u-t)\pm\sqrt{ -3u^2 -6ut-3t^2} }{2}
= \dfrac{-(u-t)\pm\sqrt{ -3 }(u+t) }{2}
= \dfrac{-(u-t)\pm i\sqrt{3 }(u+t) }{2}
=-\dfrac{u}{2}+\dfrac{t}{2}\pm\Big( \dfrac{i\sqrt{3}u }{2}+\dfrac{i\sqrt{3}t }{2} \Big)
=(-\dfrac{1}{2}\pm\dfrac{i\sqrt{3} }{2} )u +(\dfrac{1}{2}\pm\dfrac{i\sqrt{3} }{2} )t
Since u =\sqrt[3]{R} , t = -\sqrt[3]{\overline{R}}, we get two solutions with complex numbers
x = (-\dfrac{1}{2}\pm\dfrac{i\sqrt{3} }{2} )\sqrt[3]{R} +(-\dfrac{1}{2}\mp\dfrac{i\sqrt{3} }{2} )\sqrt[3]{\overline{R}}
Don't forget another known solution
x = u-t = \sqrt[3]{R}+\sqrt[3]{\overline{R}}
Finally, the same three solutions are found as using Vieta's Substitution method, which are
\begin{cases} \sqrt[3]{R}+ \sqrt[3]{\overline{R} } & \\ ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} } \\ \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R} } \end{cases}
Summary
In summary, three methods are used in the article to to explore the solutions of a depressed cubic equation. From the classic Cardano method, Vieta's Substitution and factorization method, we have to inevitably deal with the complex cubic root expressions. However, it makes the process a lot easy to understand the relationship of these expressions. In order to obtain the complete solutions, it's imperative to consider the solutions in the broad complex plane. In the end, the complete formula for the solution of a depressed cubic equation is given below.
x =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}
in which,
ω = \dfrac{-1+i\sqrt{3}}{2} and \overline{ω} =\dfrac{-1-i\sqrt{3}}{2}
The root formula looks complex but it does not necessarily mean you have to use it in solving every cubic equation. If you'd like to practice to solve a cubic equation in the real world, you may have a test in the cubic equation solver. It's a good exercise to compare solutions of different methods and enhance understandings.
Steven Zheng
posted 2 years ago