Solve the cubic equation:

$$3x^3-6x^2+4x+1=0 $$

Quick Answer

Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots.

$$ \Delta=0.099108367626886$$

$$\begin{cases} x_1=-\dfrac{\sqrt[3]{17}}{3}+\dfrac{2}{3} \\ x_2=\dfrac{\sqrt[3]{17}}{6}+\dfrac{\sqrt[3]{17}}{6}\sqrt{3}i+\dfrac{2}{3} \\ x_3=\dfrac{\sqrt[3]{17}}{6}-\dfrac{\sqrt[3]{17}}{6}\sqrt{3}i+\dfrac{2}{3} \end{cases}$$

In decimals,

$$\begin{cases} x_1=-0.19042719688608 \\ x_2=0.83333333333333+0.28867513459481i \\ x_3=0.83333333333333-0.28867513459481i \end{cases}$$

Detailed Steps on Solution

1. Convert to depressed cubic equation

The idea is to convert general form of cubic equation

$$ax^3+bx^2+cx+d = 0$$

to the form without quadratic term.

$$t^3+pt+q = 0$$

By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to

$$t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0 $$

Compare with the depressed cubic equation. Then,

$$p = \dfrac{3ac-b^2}{3a^2}$$

$$q = \dfrac{2b^3-9abc+27a^2d}{27a^3} $$

Substitute the values of coefficients, $$p, q$$ is obtained as

$$p = \dfrac{3\cdot 3\cdot 4-(-6)^2}{3\cdot 3^2}=0$$

$$q = \dfrac{2\cdot (-6)^3-9\cdot3\cdot (-6)\cdot 4+27\cdot 3^2\cdot1}{27\cdot 3^3}=\dfrac{17}{27}$$

Use the substitution to transform

Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.

$$t^3 +pt+q=0$$

Let $$x=t+\dfrac{2}{3}$$

The cubic equation $$3x³ - 6x² + 4x + 1=0$$ is transformed to

$$t^3 +\dfrac{17}{27}=0$$

For the equation $$t^3 +\dfrac{17}{27}$$, we have $$p=0$$ and $$q = \dfrac{17}{27}$$

Calculate the discriminant

The nature of the roots are determined by the sign of the discriminant.

$$\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(\dfrac{17}{27}\Big)^2}{4}+0\\ & =\dfrac{289}{2916}\\ & =0.099108367626886\\ \end{aligned}$$

1.1 Use the root formula directly

If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation.

$$t_{1,2,3} =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}$$

in which, $$ ω = \dfrac{-1+i\sqrt{3}}{2} $$ and $$ \overline{ω} =\dfrac{-1-i\sqrt{3}}{2}$$

Substitute the values of $$p, q$$ and $$\Delta$$ which we have calculated. Then,

$$\begin{aligned} \\t_1&=\sqrt[3]{-\dfrac{17}{54}+\sqrt{\dfrac{289}{2916}}}+\sqrt[3]{-\dfrac{17}{54}-\sqrt{\dfrac{289}{2916}}}\\ & =\sqrt[3]{-\dfrac{17}{54}+\sqrt{\dfrac{17^2}{54^2}}}+\sqrt[3]{-\dfrac{17}{54}-\sqrt{\dfrac{17^2}{54^2}}}\\ & =\sqrt[3]{-\dfrac{17}{54}+\dfrac{17}{54}}+\sqrt[3]{-\dfrac{17}{54}-\dfrac{17}{54}}\\ & =\sqrt[3]{\dfrac{-17}{2\cdot 3^3}+\dfrac{17}{2\cdot 3^3}}+\sqrt[3]{\dfrac{-17}{2\cdot 3^3}-\dfrac{17}{2\cdot 3^3}}\\ & =\dfrac{1}{3}\Big(\sqrt[3]{\dfrac{-17}{2}+\dfrac{17}{2}}+\sqrt[3]{\dfrac{-17}{2}-\dfrac{17}{2}}\Big)\\ & =-\dfrac{\sqrt[3]{17}}{3}\\ \end{aligned}$$

If we denote

$$R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$

$$\overline{R} = -\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$


$$\sqrt[3]{R} = 0$$, $$\sqrt[3]{\overline{R}} =-\dfrac{\sqrt[3]{17}}{3}$$

$$\begin{aligned} \\t_2&= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} }\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i\\ & =\dfrac{1}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{17}}{3}\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{17}}{3}\Big)\Big]i\\ & =\dfrac{\sqrt[3]{17}}{6}+\dfrac{\sqrt[3]{17}}{6}\sqrt{3}i\\ \end{aligned}$$

$$\begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R}}\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i \\ & =\dfrac{1}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{17}}{3}\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{17}}{3}\Big)\Big]i\\ & =\dfrac{\sqrt[3]{17}}{6}-\dfrac{\sqrt[3]{17}}{6}\sqrt{3}i\\ \end{aligned}$$

Roots of the general cubic equation

Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives

$$x_1 = t_1+\dfrac{2}{3}$$

$$x_2 = t_2+\dfrac{2}{3}$$

$$x_3 = t_3+\dfrac{2}{3}$$

2. Summary

In summary, we have tried the method of cubic root formula to explore the solutions of the equation. The cubic equation $$3x³ - 6x² + 4x + 1=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

$$\begin{cases} x_1=-\dfrac{\sqrt[3]{17}}{3}+\dfrac{2}{3} \\ x_2=\dfrac{\sqrt[3]{17}}{6}+\dfrac{\sqrt[3]{17}}{6}\sqrt{3}i+\dfrac{2}{3} \\ x_3=\dfrac{\sqrt[3]{17}}{6}-\dfrac{\sqrt[3]{17}}{6}\sqrt{3}i+\dfrac{2}{3} \end{cases}$$

in decimal notation,

$$\begin{cases} x_1=-0.19042719688608 \\ x_2=0.83333333333333+0.28867513459481i \\ x_3=0.83333333333333-0.28867513459481i \end{cases}$$

3. Graph for the function $$f(x) = 3x³ - 6x² + 4x + 1$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 3x³ - 6x² + 4x + 1$$ has one intersection point with the x-axis.

Scroll to Top