Find all possible solutions of x^3 + 2 = 0

[BurpHa]

Homework Statement

Find the exact solutions of $x ^ 3 + 2 = 0$

Relevant Equations

$x = \frac {-b \pm \sqrt {b ^ 2 - 4ac}} {2a}$

I actually know one way to solve,

$x ^ 3 + 2 = 0$
$x ^ 3 + \left (\sqrt[3] 2\right) ^ 3 = 0$
$\left(x + \sqrt[3] 2\right) \left(x ^ 2 - x\sqrt[3] 2 + \left(\sqrt[3] 2 \right)^2\right) =0$
$x + \sqrt[3] 2 = 0, x = -\sqrt[3] 2$
$x ^ 2 - x\sqrt[3] 2 + \left(\sqrt[3] 2\right)^2 = 0, x = \frac {\sqrt[3] 2} {2} \pm \frac {\sqrt[3] 2 \sqrt 3 } {2}i$
So, the solution set is $\{-\sqrt[3] 2, \frac {\sqrt[3] 2} {2} \pm \frac {\sqrt[3] 2 \sqrt 3 } {2}i \}$

However, I want to approach it like this,

$x ^ 3 + 2 = 0$
$x ^ 3 = -2$
$x = \sqrt[3]-2$
What could I do next? From what I see, there is only one solution in this way?

 

[PeroK]

Have you heard of the polar form of complex numbers?

Your approach is quite complicated, although you have rationalised denominators at least.

 

[PeroK]

Homework Statement:: Find the exact solutions of $x ^ 3 + 2 = 0$
Relevant Equations:: $x = \frac {-b \pm \sqrt {b ^ 2 - 4ac}} {2a}$

I actually know one way to solve,

$x ^ 3 + 2 = 0$
$x ^ 3 + \left (\sqrt[3] 2\right) ^ 3 = 0$
$\left(x + \sqrt[3] 2\right) \left(x ^ 2 - x\sqrt[3] 2 + \left(\sqrt[3] 2 \right)^2\right) =0$
$x + \sqrt[3] 2 = 0, x = -\sqrt[3] 2$
$x ^ 2 - x\sqrt[3] 2 + \left(\sqrt[3] 2\right)^2 = 0, x = \frac {\sqrt[3] 2} {2} \pm \frac {\sqrt[3] 2 \sqrt 3 } {2}i$
So, the solution set is $\{-\sqrt[3] 2, \frac {\sqrt[3] 2} {2} \pm \frac {\sqrt[3] 2 \sqrt 3 } {2}i \}$

This is correct

However, I want to approach it like this,

$x ^ 3 + 2 = 0$
$x ^ 3 = -2$
$x = \sqrt[3]-2$
What could I do next? From what I see, there is only one solution in this way?

There are still three solutions to that equation.

 

[BurpHa]

This is correct

There are still three solutions to that equation.

Could you tell me more? I've heard about polar form of complex numbers. How could polar form of complex numbers help in this case? Thank you.

 

[PeroK]

Could you tell me more? I've heard about polar form of complex numbers. How could polar form of complex numbers help in this case? Thank you.

There's a summary here of taking the nth root of any complex number:

https://www.123calculus.com/en/complex-number-nth-root-page-1-45-175.html

Paul's online maths notes is a good resource for lots of things:

https://tutorial.math.lamar.edu/Extras/ComplexPrimer/Forms.aspx

 

[BurpHa]

There's a summary here of taking the nth root of any complex number:

https://www.123calculus.com/en/complex-number-nth-root-page-1-45-175.html

Paul's online maths notes is a good resource for lots of things:

https://tutorial.math.lamar.edu/Extras/ComplexPrimer/Forms.aspx

Thank you for your help.

 

[FranzS]

As PeroK wrote, writing a complex number in polar form and using Euler's formula + de Moivre's theorem will give you the best insights and graphical intuition.

Another standard way to find the $n$-th complex roots of a real number is to write the variable $x$ in your initial equation as a generic complex number $a+ib$ and then... well, see what happens if you are interested.

 

[BurpHa]

As PeroK wrote, writing a complex number in polar form and using Euler's formula + de Moivre's theorem will give you the best insights and graphical intuition.

Another standard way to find the $n$-th complex roots of a real number is to write the variable $x$ in your initial equation as a generic complex number $a+ib$ and then... well, see what happens if you are interested.

Thank you.

 

[malawi_glenn]

Your "relevant equations" is not really relevant here...

Anyway, you found one solution which has real part = 0 and positive imaginary part.
The solutions to the equation $z^n = w$ are equally "spread" around a circle with the origin as the center and spaced $2\pi/n$.
You now have one solution $z_1 = 2^{1/3} \mathrm{i}$.
Now you can use simple geometry to find the other two solutions.
The radius of your circle will be $2^{1/3}$

 

[Mark44]

Your "relevant equations" is not really relevant here...

It's relevant after you pull the linear factor out of $x^3 + 2$ and are left with a quadratic polynomial factor.