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This problem came up when I was tutoring algebra and solving radical equations. Suppose R(x) = 0 is a consistent radical equation with rational powers and rational constants. It is known that that R(x) can be transformed into an nth degree polynominal P(x), assumed to be solvable with radicals, with roots xi, possibly complex. But some of the xis could be extraneous roots which have to be eliminated by verifying R(xi) = 0. Now proving this identity can be a challenge even for CASs, but this is a neccessary step in solving radical equations.

One approach that I have thought of is the following. Define z = R(xi) and construct an Nth degree polynominal, PP(z), with R(xi) as a root. Since R(xi) is suppose to identically 0 then the constant term of PP(z) will be identically zero and composed of rational or integer numbers only and thus potentially easily demostrated. Is this a valid approach?

Next suppose that P(x) is not solvable by radicals but by "ultraradicals". Will the above approach still work? Is there a rule for intger powers of an "ultraradical"?

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What is an "ultraradical"? –  Qiaochu Yuan May 17 '11 at 15:28
Google says that an ultraradical of $a$ is a solution of $x^5+x = a$. I think the question as it is stated is not clear enough to have a definite answer. What specific transformation are you applying to get $P$ from $R$, and what is the specific method you use to construct $PP(z)$? –  S. Carnahan May 17 '11 at 16:33
To S. Carnahan: The transformation of R(x) to P(x) is raising of rational powers to remove all non-integer powers. This is a standard technique and results in a polynominal of some degree. To construct PP(z) you simply set z = R(x1) and do the same process that was used to create P(X), i.e., remove all rational powers by raising to powers. –  clkirksey May 18 '11 at 13:57

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