# Complex solutions to $x^3 + 512 = 0$

An algebra book has the exercise

$$x^3 + 512 = 0$$

I can find the real solution easily enough with

$$x^3 = -512$$ $$\sqrt[3]{x^3} = \sqrt[3]{-512}$$ $$x = -8$$

The book also gives the complex solutions $$4 \pm 4\sqrt{3}i$$

But I don't understand how to find these answers. Having completed the chapter on complex numbers I can find square roots of negative numbers easily, but cube (or higher) roots are never explained.

• $x^3+a^3=(x+a)(x^2-ax+a^2)$
– user123641
Sep 2, 2014 at 3:00
• @Bryan thanks, that's very helpful. If you add that as an answer I'll accept it. I'm sure the other answers are good but I don't understand them. Sep 2, 2014 at 3:01
• $$512=8^3{}{}$$ Apr 23, 2018 at 23:31

In general $$x^3+a^3=(x+a)(x^2-ax+a^2)$$

You can apply this to get a linear term and a quadratic term in your problem. Find the roots of the quadratic term to get the other two roots.

If you want to solve the complex roots using quadratic formula, try below :

$$x^3 + 512 = x^3+8^3 = (x+8)(x^2-8x+64) = 0$$

I wanted to expand on what others have said as they don't explain what happens when you reach higher powers and I feel this solution is much easier.

For an equation in the form $$Z^n = x$$ You will have n solutions that will be evenly spaced around the origin.

So in this case we have: $$x^3 = -512$$ Which you correctly calculated one of the answers being: $$x = -8$$

Now that we know this first answer, we know the other 2 answers will be spaced 120 degrees (360/3 because there are 3 evenly placed answers) from this first answer.

So we can calculate these answers as follows: $$-8e^{\frac{120}{360}2\pi i} = 4-4\sqrt{3}i$$ and $$-8e^{\frac{240}{360}2\pi i} = 4+4\sqrt{3}i$$

Similarly if we have: $$x^4 = 4096$$ Our first answer can be calculated as $$x = \sqrt[4]{4096} = 8$$ And we know there will be 4 evenly spaced answers around the origin. Therefore we know they will be 90 degrees apart so the other 3 solutions will obviously be $$x = 8i,-8,-8i$$

Solutions of any equation of the form $x^3=a^3$ are $a$, $a\omega$, $a\omega^2$, where $\omega$ is a cube root of unity, and $\omega \neq 1$. Usually one picks $\omega=e^{2i\pi/3}$

Hint: your full solution is $x=-8z$ where $z^3=1$. If you write $z=re^{i\theta}$ ($r\in\mathbb{R})$, then $z^3=r^3e^{i3\theta}=r^3(\cos(3\theta)+i\sin(3\theta))$. What can you say about $r$ and $\theta$ given that $z^3=1$?