How does one show this complex expression equals a natural number?

Question

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"!)

Answer 1

Let $\,u = \left(-3 + \dfrac{10}{3^{3/2}}i\right)^{1/3}\,$ and $\,v = \left(-3 -\dfrac{10}{3^{3/2}}i\right)^{1/3}\,$ be the principal values of the cube roots. Being the principal value, $\,u\,$ is the root with the greatest real part of $\, z^3 = -3 + \dfrac{10}{3^{3/2}} i\,$.

Taking complex conjugates, $\,\overline{u}\,$ is a root of $\,z^3 = -3 - \dfrac{10}{3^{3/2}} i\,$. ​Given that the real part is invariant under complex conjugation, it follows that $\,\overline{u}\,$ is the root with the greatest real part of the conjugate equation, and therefore $\,v = \overline{u}\,$ (see also Sum of cube roots of complex conjugates).

We'll further note that $\,|u|^2 = \dfrac{7}{3}\,$ since $\,\displaystyle \left|- 3 \pm \dfrac{10}{3^{3/2}}i\right|^2 = 9 + \dfrac{100}{27} = \dfrac{343}{27} = \left(\dfrac{7}{3}\right)^3\,$.

Then the LHS of the top equality can be written as $\,a = u + \dfrac{|u|^2}{u} = u + \overline u\,$, and:

$$ \require{cancel} \begin{align} a^3 = (u+\overline{u})^3 &= u^3 + \overline{u}^3 + 3u\overline{u}(u+\overline{u}) \\ &= -3 + \cancel{\dfrac{10}{3^{3/2}}i} -3 - \cancel{\dfrac{10}{3^{3/2}}i} + 3|u|^2\,a \\ &= -6 + 7a \end{align} $$

Piecing it all together, $\,a = 2\,\text{Re}(u)\,$ is a root of $\,a^3-7a+6 = (a-2)(a-1)(a+3)\,$, and by the choice of the principal value of the cube root it is the largest real root, so in the end $\,a=2\,$.

Answer 2

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 $$ \frac{\sqrt{7}}{\sqrt{3}}\exp\left(\frac{i\pi}{3} - \frac{i}{3}\sin^{-1}\frac{10}{7\sqrt{7}}\right) = \frac{\sqrt{7}}{\sqrt{3}}\exp\left(i\sin^{-1}\frac{2}{\sqrt{7}}\right) = \frac{\sqrt{7}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{7}} + i\frac{2}{\sqrt{7}}\right) = 1+i\frac{2}{\sqrt{3}}. $$

Answer 3

My maths professor when I was a first year alumni showed us the demonstration of the cubic formula, and then told us it was pretty useless for several reasons:

• if you need a numeric approximation of the roots, the cubic formula needs one square and three cubic roots of complex numbers (which involves more cubic equation solutions). It is much faster to compute directly the roots by Newton-Raphson.
• It is when the three roots are real that all the solutions have to be expressed as sums of complex numbers.
• there are several cases were you get complicated formula for just representing an integer. There is no general method to generate this integer. There are general methods to show that the original equation has at least one integer root (it is reducible on the rationals).
• there is also a method to give roots of equations of the fourth degree as a function of radicals, but it is known that there is no general formula for equations of the fifth degree or more (Galois theory).

Note that the equation you have shown is of the form: $$x^3-(n+1)x+n=0$$ and has obviously $1$ as a solution. I wouldn't say that you have to "guess" this solution. The other solutions can be easily found after factorizing $x-1$.