Simplifying a nested radical that includes a complex number

[cbarker1]

TL;DR

Simplifying nested radical that has a complex number under it.

Dear Everyone,

This post is not a homework assignment...
I want to use the quartic formula. In one step is to solve the resolvent cubic. I know that there is 3 real solutions this particular resolvent cubic. I want to know how Bombelli got his answers before the discovery of the trigonometric method for the casus irrecidulus.
I have this radical $\sqrt[3]{\frac{24128}{27}-\frac{2520i\sqrt{3}}{27}}$. I know that if the radicand is a perfect cube, then I can cancel the cube root with the cube. So $(a-bi\sqrt{3})^3= \frac{24128}{27}-\frac{2520i\sqrt{3}}{27}$. We can equate the Real part of both $(a-bi\sqrt{3})^3$ and the $\frac{24128}{27}-\frac{2520i\sqrt{3}}{27}$ and the imagery part of both $(a-bi\sqrt{3})^3$ and the $\frac{24128}{27}-\frac{2520i\sqrt{3}}{27}$. To yield this system of polynomial equations: $$
\begin{align*}
a^3-9ab^2 &= \frac{24128}{27} \\
3b^3-3a^2b &= \frac{2520}{27} \\
\end{align*}$$

We know that $a,b\not=0$.I am having a hard time solving this system by hand. Is there any tips on solving a system of polynomials?

Thanks,
Cbarker1

 

[mfb]

If you introduce c=b/a your first equation will contain a³ and a³ c² while the second equation contains a³ c³ and a³ c. Divide both equations by each other and you get an expression that only depends on c. Unfortunately it's a cubic equation again, and I don't know if its solutions are nice.

 

[cbarker1]

If you introduce c=b/a your first equation will contain a³ and a³ c² while the second equation contains a³ c³ and a³ c. Divide both equations by each other and you get an expression that only depends on c. Unfortunately it's a cubic equation again, and I don't know if its solutions are nice.

I looked the solution on wolframalpha where I will type the solution form as $(a,b)$. $(-\frac{16}{3},\frac{14}{3})$, $(\frac{29}{3},\frac{1}{3})$, and $(\frac{-13}{3},-5)$. It has nice solutions...

 

[Office_Shredder]

Factoring it actually looks nice, you get $a(a-3b)(a+3b)$ and (after dividing by 3) $b(b-a)(b+a)$.

I don't really know where to go from there.

 

[Frabjous]

note: I did this by hand, and I have a sign difference with the wolframalpha solution, but I think the observation is correct
I factored out 2/3 at the beginning giving
$3016c^3+945c^2-3016c-105 = 0$
If you look at the factors of 3016 (f) and -105 (g), you can by guesswork find a solution of the form
$(f_1c-g_1)(f_2c-g_2)(f_3c-g_3)$
which works because there are rational solutions of 7/8, -1/29 and -15/13

 

[stevendaryl]

In general, taking cube roots of complex numbers looks like this:

$(a + bi)^3 = c + di$

With the benefit of hindsight, we can redefine:

$R = \sqrt{a^2 + b^2}$
$x = a/R$

$S = \sqrt{c^2 + d^2}$
$y = c/S$

So the problem becomes to find $R$ and $x$ in terms of $S$ and $y$.

The first part is easy: $R = S^{1/4}$

Then the equation for $x$ becomes

$(x +i \sqrt{1-x^2})^3 = y +i\sqrt{1-y^2}$

Cubing the left sude and taking just the real part gives:

$x^3 - 3x(1-x^2)= y$
or
$4x^3 -3x = y$

So this reduces the problem to solving a real cubic equation. Not sure that’s an improvement...