Solve the cubic equation:

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

Quick Answer

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

$$ \Delta=0.5625$$

$$\begin{cases} x_1=-\sqrt[3]{\dfrac{3}{2}}+1 \\ x_2=\dfrac{1}{2}\sqrt[3]{\dfrac{3}{2}}+\dfrac{1}{2}\sqrt[3]{\dfrac{3}{2}}\sqrt{3}i+1 \\ x_3=\dfrac{1}{2}\sqrt[3]{\dfrac{3}{2}}-\dfrac{1}{2}\sqrt[3]{\dfrac{3}{2}}\sqrt{3}i+1 \end{cases}$$

In decimals,

$$\begin{cases} x_1=-0.14471424255333 \\ x_2=1.5+0.86602540378444i \\ x_3=1.5-0.86602540378444i \end{cases}$$

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

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

$$q = \dfrac{2\cdot (-6)^3-9\cdot2\cdot (-6)\cdot 6+27\cdot 2^2\cdot1}{27\cdot 2^3}=\dfrac{3}{2}$$

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 $$x=t+1$$

The cubic equation $$2x³ - 6x² + 6x + 1=0$$ is transformed to

$$t^3 +\dfrac{3}{2}=0$$

For the equation $$t^3 +\dfrac{3}{2}$$, we have $$p=0$$ and $$q = \dfrac{3}{2}$$

Calculate the discriminant

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

$$\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(\dfrac{3}{2}\Big)^2}{4}+0\\ & =\dfrac{9}{16}\\ & =0.5625\\ \end{aligned}$$

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,

$$\begin{aligned} \\t_1&=\sqrt[3]{-\dfrac{3}{4}+\sqrt{\dfrac{9}{16}}}+\sqrt[3]{-\dfrac{3}{4}-\sqrt{\dfrac{9}{16}}}\\ & =\sqrt[3]{-\dfrac{3}{4}+\sqrt{\dfrac{3^2}{4^2}}}+\sqrt[3]{-\dfrac{3}{4}-\sqrt{\dfrac{3^2}{4^2}}}\\ & =\sqrt[3]{-\dfrac{3}{4}+\dfrac{3}{4}}+\sqrt[3]{-\dfrac{3}{4}-\dfrac{3}{4}}\\ & =-\sqrt[3]{\dfrac{3}{2}}\\ \end{aligned}$$

If we denote

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

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

then,

$$\sqrt[3]{R} = 0$$, $$\sqrt[3]{\overline{R}} =-\sqrt[3]{\dfrac{3}{2}}$$

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

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

Roots of the general cubic equation

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

$$x_1 = t_1+1$$

$$x_2 = t_2+1$$

$$x_3 = t_3+1$$

2. Summary

In summary, we have tried the method of cubic root formula to explore the solutions of the equation. The cubic equation $$2x³ - 6x² + 6x + 1=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

in decimal notation,