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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.

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    $\begingroup$ $x^3+a^3=(x+a)(x^2-ax+a^2)$ $\endgroup$
    – user123641
    Sep 2, 2014 at 3:00
  • $\begingroup$ @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. $\endgroup$
    – friedo
    Sep 2, 2014 at 3:01
  • $\begingroup$ $$512=8^3{}{}$$ $\endgroup$
    – Mr Pie
    Apr 23, 2018 at 23:31

5 Answers 5

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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.

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

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

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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 $$x = 8i,-8,-8i$$

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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 $\omega=e^{2i\pi/3}$

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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 $z^3=1$?

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