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How do you rationalize the denominator of something like $$\frac{1}{\sqrt[n]{a_1}+\sqrt[n]{a_2}+...+\sqrt[n]{a_n}}$$? I'm thinking roots of unity.

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  • $\begingroup$ The edit seems to have changed the question, @Aryabhata. $\endgroup$ – Lubin Jan 4 '15 at 21:27
  • $\begingroup$ @Lubin: No, I just changed inline math to centered on its own line. No other changes. $\endgroup$ – Aryabhata Jan 4 '15 at 22:19
  • $\begingroup$ See the end of my answer to Rationalizing radicals. $\endgroup$ – Dave L. Renfro Jan 5 '15 at 20:22
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Hint: you should rationalize step by step for each variable by using the simple reason $$ \frac{1}{\sqrt[n]{x}+y}=\frac{f(x,y)}{x \pm y^n}, $$ where $f(x,y)$ is the suitable expression such that $(\sqrt[n]{x}+y)f(x,y)=x\pm y^n$.

Example. Let us rationalize $\dfrac{1}{\sqrt[3]{a_1}+\sqrt[3]{a_2}+\sqrt[3]{a_3}}$. Put $\sqrt[3]{a_2}+\sqrt[3]{a_3}=y$ we have
$$ \dfrac{1}{\sqrt[3]{a_1}+y}=\frac{{y}^{2}-y\sqrt [3]{a_{{1}}}+{a_{{1}}}^{2/3}}{(a_1+y)({y}^{2}-y\sqrt [3]{a_{{1}}}+{a_{{1}}}^{2/3})}=\frac{{y}^{2}-y\sqrt [3]{a_{{1}}}+{a_{{1}}}^{2/3}}{(a_1+y)({y}^{2}-y\sqrt [3]{a_{{1}}}+{a_{{1}}}^{2/3})}=\frac{{y}^{2}-y\sqrt [3]{a_{{1}}}+{a_{{1}}}^{2/3}}{(a_1+y^3)}=\frac{{y}^{2}-y\sqrt [3]{a_{{1}}}+{a_{{1}}}^{2/3}}{a_1+a_{{2}}+a_3+3\,\sqrt [3]{a_2^2 a_3}+3\,\sqrt [3]{a_{{2}}a_3^2}}. $$ So,we rationalize the expression with respect to the variable $a_1.$ Then will repeat it for $a_2$ and $a_3.$

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  • $\begingroup$ I have added a sample $\endgroup$ – Leox Jan 4 '15 at 21:49
  • $\begingroup$ I dont prove such statement. Seems it can be done by induction. $\endgroup$ – Leox Jan 4 '15 at 22:51
  • $\begingroup$ There is a general approuch which use the symmetric functions $\endgroup$ – Leox Jan 4 '15 at 23:22
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Let’s take a couple of very simple examples. The first one is your familiar $1/(a+\sqrt b)$, where, as you know, you multiply by the conjugate expression of the denominator, but the result downstairs is the norm of the denominator from its field down to the rational field.

For a more complicated example, take $1/(a + \root3\of b)$. Here, the norm is $$ (a+\root3\of b)(a+\omega\root3\of b)(a+\omega^2\root3\of b)\,, $$ where $\omega=(-1+\sqrt{-3})/2$, a primitive cube root of unity. And you get the rationalized form of your fraction by multiplying top and bottom by the two factors of the norm that don’t originally appear. The final result is $$ \frac{a^2-a\root3\of b +\root3\of b^2}{a^3+b}\,. $$ You would use the same technique for rationalizing, say, $1/(a+b\root3\of5+c\root3\of5^2)$.

You have asked about a denominator that’s rather more complicated, of course; call it $D$, and call $K$ the field generated by the separate monomials in $D$, as extension of $\Bbb Q$. If you were talking about roots of unity, that would be a normal extension of $\Bbb Q$, but in the general situation, $K$ would be nonnormal, and you’d need to go to the Galois closure (a splitting field) over $\Bbb Q$, call it $L$. Then the conjugates of the denominator will all lie in $L$, and you need to find them all and multiply by their product.

And now I have to tell you that there’s a simpler way, but still a computational mess, a way that doesn’t involve dealing with the Galois closure of $K$, but just works in $K$ itself. You use the regular representation of $K$ as $n$-by-$n$ matrices over $\Bbb Q$, where $n=[K\colon\Bbb Q]$, so that every element of $K$ can be regarded as a matrix. You want the matrix-inverse of the matrix representing $D$, the denominator of your fraction. Then you see that the top of your desired fraction is the adjugate matrix of this, but written as a polynomial expression in the irrationalities appearing in $D$, which seems to me to be a mess to do; and the bottom is just the determinant of the matrix of $D$, which in fact is just the field-theoretic norm that I mentioned before.

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