<s>[INST] How does one show this complex expression equals a natural number? We have:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3}+ \frac{7}{3 \left(\frac{10}{3^{3/2}}i-3\right)^{1/3}}=2$$
This comes from solving the cubic equation of $x^3-7x+6=0$ which factors as $(x-2)(x-1)(x+3)=0$
We can simplify the problem into finding just one part of this, namely:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3} = 1 +  \frac{2}{\sqrt{3}} i$$
Now, is there any general method (which doesn't involve factorising a cubic!) in which we get from the LHS of the above equation to the RHS? Or do we simply say that the cubic formula fails in this case and we have to resort to trial-and-error factorisation to find the result?
In which case, if there is a general method, could we not express this method as another cubic formula circumvents the intermediate complex steps?
BTW, I am only interested in cases in which the solutions of the cubic are rational or real algebraic solutions (without complex sub-parts). So we might be able to use this fact in a general method.
(Some might say that even finding $27^{1/3}$ is trial-and-error in a way, since we could try numbers 1,2,3... to see which one works. But we shall ignore and just say that finding the cube root of an integer is "allowed"!)
 [/INST] Write your complex numbers in polar. Then you have
$$
\left(-3+i\frac{10}{3\sqrt{3}}\right)^{1/3}=\left[\frac{7\sqrt{7}}{3\sqrt{3}}\exp\left(i\pi-i\sin^{-1}\frac{10}{7\sqrt{7}}\right)\right]^{1/3} = \frac{\sqrt{7}}{\sqrt{3}}\exp\left(\frac{i\pi}{3} - \frac{i}{3}\sin^{-1}\frac{10}{7\sqrt{7}}\right).$$
Now honestly simplifying that $\sin^{-1}$ would require solving a cubic equation. However, since we know our answer should come out to $\pi/3 - \sin^{-1}(2/\sqrt{7}) = \sin^{-1}(1/\sqrt{28})$, we can cheat and calculate
$$
\sin\left(3\sin^{-1}\frac{1}{2\sqrt{7}}\right) = \frac{3}{2\sqrt{7}}-\frac{4}{56\sqrt{7}} = \frac{10}{7\sqrt{7}},
$$
allowing us to miraculously guess the answer. Thus, $\pi/3 - \sin^{-1}(10/\sqrt{343})/3 = \sin^{-1}(2/\sqrt{7})$, and we have
$$
\frac{\sqrt{7}}{\sqrt{3}}\exp\left(\frac{i\pi}{3} - \frac{i}{3}\sin^{-1}\frac{10}{7\sqrt{7}}\right) = \frac{\sqrt{7}}{\sqrt{3}}\exp\left(i\sin^{-1}\frac{2}{\sqrt{7}}\right) = \frac{\sqrt{7}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{7}} + i\frac{2}{\sqrt{7}}\right) = 1+i\frac{2}{\sqrt{3}}.
$$
</s>