# Cubic roots

1. Jul 13, 2007

### edgo

Hi! Who knows:
can any cubic root like $\sqrt[3]{x}$ with x real be written as a form in which only square roots (real or complex) are involved?

2. Jul 13, 2007

### edgo

Sorry, not any but every!!!

3. Jul 13, 2007

### gunch

Yes it can, but sometimes the square root have to be of an irrational number like another cubic root so while it's the square root of a real number we haven't gotten rid of cubic roots. For example we have: $$\sqrt[3]{2}=\sqrt{\sqrt[3]{4}}$$ or $$\sqrt[3]{-2}=\sqrt{\sqrt[3]{4}i}$$.

If x is nonnegative then we have a nonnegative real $$y=\sqrt[3]{x^2}$$. Clearly $$x^2$$ exists for all real x and is itself a real nonnegative number. Similarly every nonnegative real number have a nonnegative real cubic root, so $$\sqrt[3]{x^2}$$ also exists and is a nonnegative real. The square root of y also exist since all real nonnegative numbers have a real nonnegative square root.
$$\sqrt{y}=\sqrt[2*3]{x^2}=\sqrt[3]{x}$$
So the square root of y exists for all nonnegative x and it's equal to the cubic root of x.

For negative x we have a positive real number z such that:
x = -1z
So:
$$\sqrt[3]{x}=\sqrt[3]{-1z}=\sqrt[3]{-1}\sqrt{z}=-\sqrt[3]{z}$$
Since z is a nonnegative real number we know that we can write its cubic root as the square root of a real number. Let $$\sqrt{y}=\sqrt[3]{z}$$, then we have:
$$-\sqrt[3]{z} = \sqrt{i} \sqrt{y} = \sqrt{yi}$$
So for all negative x we can write its cubic root as the square root of a pure imaginary number.

4. Jul 14, 2007

### HallsofIvy

Staff Emeritus
Yes, obviously, if a is any real number, I can define y to be a2 and then say $a= \sqrt{y}$. I don't think that was what edgo meant.

In general, a cube root cannot be written in a form involving only square roots and rational numbers. That is because any number written in that form must be algebraic of order a power of two while any cuberoot is obviously algebraic of order 3.

5. Jul 14, 2007

### mathwonk

the proof halls is giving depends on the fact that there is a certain vector space associated to the root, and the dimension of that vector space equals the degree of the polynomial, namely 3. Moreover, if the expresion is writtena s a composition, the degrees are multiplicative, so if a square root were involved the product would be even, a contradiction. this is elementary field theory, and is based on the fact that the k vector space

k[X]/(f(X)) where f is a polynomial of degree d, has k dimension d.

6. Jul 14, 2007

### edgo

Three times: Thank you.
I'm not (yet?) familiar with the Galois theory. Then, I assumed that the cubic roots which are involved in the solution - via the Cardano or Harriot algorithms - of cubic equations were there because the algorithm introduced them. This not being so has the consequence that in the value of cos x etc. in general cubic roots are involved (since you can also solve most cubic equations with a goniometric algorithm). Am I right?

7. Jul 15, 2007

### edgo

There should not be such a thing as a quick reply in math. Rereading my last remarks urges me to give a correction on "you can solve most cubic equations with a goniometric algorithm". I meant cubic equations with 3 real roots and furthermore, most makes no sense at all as we are talking about the set of cubic equations and that set doesn't have a limited number of elements.

8. Jul 15, 2007

### HallsofIvy

Staff Emeritus
It didn't occur to you that it might be because solving an equation is an "inverse" problem and that the inverse of cubing is the cube root?

I don't know what you mean "goniometric algorithm". A google search for "goniometric algorithm" turn up nothing and a google search for "goniometric" led to one reference giving "trigonometric" as a secondary definition. All other references were to measurement devices.

9. Jul 15, 2007

### edgo

You know this goniometric algorithm for sure!
We reduce any cubic equation (further: CE) with real coefficients to a form $V=x^3+px+q=0$. Then V=0 also has real coefficients.
When the discriminant $D=\frac{q^2}{4}+\frac{p^3}{27}\prec0$ we have the “casus irreducibilis”, in which case V=0 has three real roots.

We set $x=r\cos\alpha$ and substitute that x into V=0: $r^3{(\cos\alpha)}^3+pr\cos\alpha+q=0$.
There exists a goniometric equality $\cos3\alpha=4{(\cos\alpha)}^3-3\cos\alpha$.
So, we multiply the terms of the left side of V=0 by 4 and replace $4{(\cos\alpha)}^3$ by $\cos3\alpha+3\cos\alpha$.
It gives
$r^3(\cos3\alpha+3\cos\alpha)+4pr\cos\alpha+4q=0$
then
$r^3\cos3\alpha+4q+(3r^2+4p)rcos\alpha=0$
We can choose $3r^2+4p=0$ and this makes $r^3\cos3\alpha+4q=0$

We assume that $\beta$ meets the condition
$\alpha=\beta frac{2k\pi}{3}$

The roots of $V_1=0$ can be written now as $x_k=r\cos(\beta+ \frac{2k\pi}{3})$ with k=1,2,3.

I owe this solution to Dr P. Wijdenes; it’s dated from the times of the pre-calculator math (not to be confused with the pre-calculus math).

No it didn't occur to me that solving a CE is the inverse of cubing. If I have a polynom of the 3rd power with real coefficients and I multiply this polynom
with a factor, say, x-2, then I have a quartic equation with three cubic rootsand a root x=2.
All I want to say is that it's not at the first sight clear: there are dark corners in the set of real numbers.

10. Jul 15, 2007

### edgo

Sorry, again a correction.
False is: we assume that $\beta$ meets the condition etc.
Correct is:
We have to solve $\alpha$ in$cos3\alpha=frac{-4p}{r^3}$

We assume that $\beta$ is a solution for $\alpha$ in this equation.

then $\alpha=\beta+frac{2k\pi}{3}$.

Now the roots $x_k$ of V=0 are equal to $rcos(\beta+frac{2k\pi}{3})$ with k=1,2,3.

11. Jul 15, 2007

### edgo

Sorry, again a correction.Now I understand what went wrong!
False is: we assume that $\beta$ meets the condition etc.
Correct is:
We have to solve $\alpha$ in $cos3\alpha=\frac{-4p}{r^3}$

We assume that $\beta$ is a solution for $\alpha$ in this equation.

then $\alpha=\beta+\frac{2k\pi}{3}$.

Now the roots $x_k$ of V=0 are equal to $rcos(\beta+\frac{2k\pi}{3})$ with k=1,2,3.