## Cubic roots

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?
 Sorry, not any but every!!!
 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.

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## Cubic roots

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.
 Recognitions: Homework Help Science Advisor 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.
 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?
 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.

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 Quote by 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.
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?

 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?
 Quote by 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.
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.
 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.
 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.
 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.