I Simplifying a nested radical that includes a complex number

AI Thread Summary
The discussion revolves around simplifying a nested radical involving a complex number, specifically the expression ##\sqrt[3]{\frac{24128}{27}-\frac{2520i\sqrt{3}}{27}##. The user seeks to understand how to solve a system of polynomial equations derived from equating the real and imaginary parts of the expression, leading to two equations involving variables a and b. Suggestions include substituting c = b/a to reduce the system, although this results in another cubic equation. The user also mentions finding solutions using WolframAlpha, which yields rational solutions, and discusses the general approach to taking cube roots of complex numbers. The conversation highlights the complexity of solving such equations and the potential for rational solutions.
cbarker1
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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
 
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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.
 
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mfb said:
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...
 
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.
 
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
 
Last edited:
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...
 
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