Finding rational roots to an equation (univariate) [closed]

So I'm using Mathematica 7 for Students and I'm using the Root function to find the roots of an equation over the rational numbers:

$$225x^6 -343 = 0$$

I created this equation from trying to prove that the number $\frac{\sqrt7}{\sqrt[3]15}$ is irrational or rational, and using the rational roots theorem, I came up with 24 possible roots that I don't want to test out manually. Using Root gives me the following output for some reason:

Root[225*x^6 - 343 == 0, x]
Root::nup: -343 + 225x^5 is not a univariate polynomial >>


How is this polynomial not univariate? I'm pretty sure it is, since it only has $x$ in it. Also, another gem occurs when I try the same thing with $x^2 = 1$:

Root[x^2 - 1 == 0, x]
Root::nup: x^2 - 1 == 0 is not a univariate polynomial >>


What is going on?

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Use Solve, not Root. Root is mainly meant for representing solutions, not for computing them, and the syntax you used for Root is not correct (see the docs) – Szabolcs Mar 9 at 17:15
as Szabolcs said, look at the docs. you can do this by placing your cursor on Root and pressing F1 (in particular, look at the examples section). Quick and easy (in fact faster than asking here). – acl Mar 9 at 17:30
Possibly you are confusing Root for the function Roots. – Daniel Lichtblau Mar 9 at 20:07

closed as too localized by Artes, Sjoerd C. de Vries, whuber, m_goldberg, Yves KlettMar 10 at 16:06

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Use Solve, not Root.

Solve[225*x^6 - 343 == 0, x]

(* ==>
{{x -> -(-(1/15))^(1/3) Sqrt[7]}, {x -> (-(1/15))^(1/3) Sqrt[
7]}, {x -> -(Sqrt[7]/15^(1/3))}, {x -> Sqrt[7]/15^(
1/3)}, {x -> -(((-1)^(2/3) Sqrt[7])/15^(1/3))}, {x -> ((-1)^(2/3)
Sqrt[7])/15^(1/3)}}
*)


Root is for representing roots of polynomials, not for computing them. Also, the syntax you used is simply incorrect (please check the docs for Root). The correct syntax to represent the 1st root of this equation would be:

 Root[225*#^6 - 343 &, 1]


You'll notice that this doesn't give a form in radicals automatically, it just returns unevaluated. Use ToRadicals to do the conversion.

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