How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

In polar form, 
$$x = 35+18 i \sqrt{3} = \sqrt{13}^3 \cdot \exp\left( i \cdot \rm{atan} \frac{18\sqrt{3}}{35}\right)$$
so that
$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \; \cos\left(\frac{1}{3} \rm{atan} \frac{18\sqrt{3}}{35} \right)$$
which magically evaluates to $7$ in a calculator.
(I know, I know: It's not appropriate to call any one value the cube root of another. Be that as it may ...)
To verify that the calculation is exact, we'll seek $T = \tan\theta$ such that $\tan 3\theta = 18\sqrt{3}/35$.
Invoking the triple-angle tangent formula, we have
$$\tan3\theta = \frac{T ( 3 - T^2 )}{1 - 3 T^2} = \frac{18\sqrt{3}}{35}$$
If I were convinced that $T$ is a relatively nice number, I'd be willing to believe that the $T$ in the left-hand numerator accounts for the $\sqrt{3}$ in the right-hand numerator. Write $T := S \sqrt{3}$, so that:
$$\frac{S \; ( 1 - S^2 )}{1 - 9 S^2 } = \frac{6}{35}$$
Now, note that (negative) $35$ is just one away from (negative) $36$, a perfect square (and multiple of $9$, to boot), so that comparing denominators suggests $S = \pm 2$. By happy coincidence, $S=2$ gives precisely the numerator we'd need.
Therefore, $\tan\theta = T = S\sqrt{3} = 2 \sqrt{3}$, so that $\cos\theta = \frac{1}{\sqrt{1 + T^2}} = \frac{1}{\sqrt{13}}$. This gives
$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 2$$
and we're don--- hey, waydaminnit ... $2\neq 7$! What happened?
Well, notice that we actually solved
$$\frac{S(1-S^2)}{1-9S^2} = \frac{-6}{-35} = \tan(3\theta+\pi)$$
("Wait, what's the difference?" "Just keep reading.") so that we actually got
$$T = \tan(\theta+\pi/3) = 2\sqrt{3}$$
That is, what we took for $\theta$ was in fact $\theta+\pi/3$. We can get at the $\theta$ we wanted thusly:
$$\tan\theta = \tan\left((\theta+\pi/3)-\pi/3\right) = \frac{T-\tan\frac{\pi}{3}}{1+T\tan\frac{\pi}{3}} = \frac{2\sqrt{3}-\sqrt{3}}{1+2\sqrt{3}\cdot\sqrt{3}} = \frac{\sqrt{3}}{7}$$
so that $\cos\theta = \frac{7}{2\sqrt{13}}$ and therefore $\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 7$, as expected.

So, why the $2$ before?
When we solved using $\frac{-6}{-35}$ instead of $\frac{6}{35}$, we implicitly replaced $x$ ---a complex number having positive real and imaginary components--- with $-x$. (Recall that $\tan 3\theta$ is the ratio of these components. It can't distinguish $x$ from $-x$.) Consequently, the $T$ we first obtained corresponds to the relation:
$$\sqrt[3]{-x} = \sqrt{13} \left( \frac{1}{\sqrt{13}} + i \frac{2\sqrt{3}} {\sqrt{13}} \right) = 1 + 2 i \sqrt{3}$$
which is to say: $(1+2i\sqrt{3})^3 = - \left( 35 + 18 i \sqrt{3}\right)$.
Then, since the arguments of $x$ and $-x$ differ by $\pi$, the arguments of the calculated cube roots must differ by $\pi/3$, and our second computation gives $\theta$ corresponding to this result:
$$\sqrt[3]{x} = \sqrt{13} \left( \frac{7}{2\sqrt{13}} + i \frac{\sqrt{3}}{2\sqrt{13}}\right) = \frac{1}{2}\left( 7 + i \sqrt{3} \right)$$
