# Simplifying Ugly Radical Expressions from Cardano's Method

• Jul 31st 2007, 05:44 AM
topsquark
Simplifying Ugly Radical Expressions from Cardano's Method
I asked a similar question to this some time ago, but as we have gained several new specialists since then, I thought I'd try again.

The problem comes from solving the equation
$x^3 - 7x + 6 = 0$
by Cardano's method.

I get an expression for x of the form:
$x = 2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right )$

As it happens,
$2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right ) = 3$

but how can I prove that? Is there any way to simplify this monster without knowing the answer ahead of time?

-Dan
• Jul 31st 2007, 07:32 AM
ThePerfectHacker
I am not sure what you are asking.
But sometimes I get expressions which look like,

$\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}$.

And in general there is no way to simplify it, you just leave it like that. Part of the difficulty in the word "simplify" is that it is not well-defined, and hence we cannot come up with a theory which tells us when we can "simplify" and when we cannot. Perhaps you can create a definition.

And, I also do not understand how you got trigonometric functions in your answer. I never remember learning it that.
• Jul 31st 2007, 08:14 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I am not sure what you are asking.
But sometimes I get expressions which look like,

$\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}$.

And in general there is no way to simplify it, you just leave it like that. Part of the difficulty in the word "simplify" is that it is not well-defined, and hence we cannot come up with a theory which tells us when we can "simplify" and when we cannot. Perhaps you can create a definition.

And, I also do not understand how you got trigonometric functions in your answer. I never remember learning it that.

The answer is that I would like to see if there is a way I can know that
$2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right ) = 3$
without resorting to a decimal approximation.

I did make a mistake in my derivation. (That's what I get for not writing it out in full.)

As to how I got that form:
$x^3 - 7x + 6 = 0$

Let $x = p - q$ and $q = -\frac{7}{3p}$

Then

$p^3 + \frac{343}{27p^3} + 6 = 0$

$27p^6 + 162p^3 + 343 = 0$

$p^3 = -3 + i \frac{10\sqrt{3}}{9}$ is one solution.

Now put $p^3$ into complex polar form. (This will "simplify" the eventual answer for x.)

$p^3 = \frac{7\sqrt{21}}{9} \cdot e^{i \cdot atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{ \pi}{2}}$

So
$p = \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}}$

Thus
$q = -\frac{7}{3p} = -\frac{7}{3} \sqrt[3]{\frac{9}{7\sqrt{21}}}
e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}$

$q = -\sqrt[3]{\frac{7\sqrt{21}}{9}} e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}$

Thus
$x = p - q = \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}} + \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}$

$x = 2 \sqrt[3]{\frac{7\sqrt{21}}{9}} \left ( \frac{ e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}} +
e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}}{2} \right )$
<-- This is the reason for the complex polar form

$x = 2 \sqrt[3]{\frac{7\sqrt{21}}{9}} cos \left (
\frac{1}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + \frac{\pi}{6} \right )$

$x = 2$

-Dan