Solve the cubic equation:
$$x^3+3=0 $$
Quick Answer
Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots.
$$ \Delta=2.25$$
$$\begin{cases} x_1=-\sqrt[3]{3} \\ x_2=\dfrac{\sqrt[3]{3}}{2}+\dfrac{\sqrt[3]{3}}{2}\sqrt{3}i \\ x_3=\dfrac{\sqrt[3]{3}}{2}-\dfrac{\sqrt[3]{3}}{2}\sqrt{3}i \end{cases}$$
In decimals,
$$\begin{cases} x_1=-1.4422495703074 \\ x_2=0.5+0.86602540378444i \\ x_3=0.5-0.86602540378444i \end{cases}$$
Detailed Steps on Solution
1. Use the sum of cubes identity
This cubic equation $$x³ + 3=0$$ contains only a cubic term and a constant term. It might be the simplest form of a cubic equation.
Isolate the leading term by moving the constant term to the right hand side.
$$x^3 = -3$$
and then take cubic roots on the both sides, which results in
$$x =\sqrt{3}$$
If the solution is to find only real root, that's all. However, if it is to find all solutions, need more steps since a cubic eqaution normally has three roots.(Some may have duplicate ones)
Factorization by sum of cubes identity
To find complete roots, use the sum of cubes formula to factorize the binomial
$$\Big(x+\sqrt{3}\Big)\Big[(x)^2-x\cdot \sqrt{3}+(\sqrt{3})^2\Big] = 0$$
Now the binomial on the left hand side is transformed to a product of two expressions. If the above equation is to hold true, either of the expressions is equal to zero.
Then we get a linear equation
$$x+\sqrt{3}=0$$
and a quadratic equation.
$$(x)^2-x\cdot \sqrt{3}+(\sqrt{3})^2=0$$
Solve the linear equation $$x+\sqrt{3}=0$$
Subtract the value of the constant term from both sides yields the first solution.
$$x_1=-\sqrt[3]{3}\approx-1.4422495703074$$
Solve the quadratic equation $$(x)^2-x\cdot \sqrt{3}+(\sqrt{3})^2$$
Let $$m = -\sqrt[3]{3}$$, then the equation is simplified as
$$x^2+mx+m^2= 0$$
Using the root identity for a quadratic equation, $$x$$ is found to be the product of $$m$$ and a complex number.
$$\begin{aligned}\\ x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} \\ &=\dfrac{-m\pm m\sqrt{3}i}{2} \\ \end{aligned}$$
Substitute the value of $$m$$, that is, $$m=-\sqrt[3]{3}$$, then we get another two complex roots
$$x_2 =\dfrac{\sqrt[3]{3}}{2}\Big(1+ \sqrt{3}i\Big)$$
$$x_3 =\dfrac{\sqrt[3]{3}}{2}\Big(1- \sqrt{3}i\Big)$$
2. Cubic Root Formula
The equation $$x³ + 3=0$$ has no quadratic term, compared with the general cubic equation. We can use the root formula to calculate the roots direvtly.
$$t^3 +pt+q=0$$
For the equation $$$$, we have $$p=0$$ and $$q = 3$$
Calculate the discriminant
The nature of the roots are determined by the sign of the discriminant.
$$\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{3^2}{4}+0\\ & =\dfrac{9}{4}\\ & =2.25\\ \end{aligned}$$
2.1 Use the root formula directly
If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation.
$$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}{2}+\sqrt{\dfrac{9}{4}}}+\sqrt[3]{-\dfrac{3}{2}-\sqrt{\dfrac{9}{4}}}\\ & =\sqrt[3]{-\dfrac{3}{2}+\sqrt{\dfrac{3^2}{2^2}}}+\sqrt[3]{-\dfrac{3}{2}-\sqrt{\dfrac{3^2}{2^2}}}\\ & =\sqrt[3]{-\dfrac{3}{2}+\dfrac{3}{2}}+\sqrt[3]{-\dfrac{3}{2}-\dfrac{3}{2}}\\ & =-\sqrt[3]{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} }$$
then,
$$\sqrt[3]{R} = 0$$, $$\sqrt[3]{\overline{R}} =-\sqrt[3]{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(-\sqrt[3]{3}\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\sqrt[3]{3}\Big)\Big]i\\ & =\dfrac{\sqrt[3]{3}}{2}+\dfrac{\sqrt[3]{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]{3}\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\sqrt[3]{3}\Big)\Big]i\\ & =\dfrac{\sqrt[3]{3}}{2}-\dfrac{\sqrt[3]{3}}{2}\sqrt{3}i\\ \end{aligned}$$
3. Summary
In summary, we have tried the method of to explore the solutions of the equation. The cubic equation $$x³ + 3=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.
$$\begin{cases} x_1=-\sqrt[3]{3} \\ x_2=\dfrac{\sqrt[3]{3}}{2}+\dfrac{\sqrt[3]{3}}{2}\sqrt{3}i \\ x_3=\dfrac{\sqrt[3]{3}}{2}-\dfrac{\sqrt[3]{3}}{2}\sqrt{3}i \end{cases}$$
in decimal notation,
$$\begin{cases} x_1=-1.4422495703074 \\ x_2=0.5+0.86602540378444i \\ x_3=0.5-0.86602540378444i \end{cases}$$
4. Graph for the function $$f(x) = x³ + 3$$
Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x³ + 3$$ has one intersection point with the x-axis.