How would you rationalize a denomiator with $3$ or more arbitary roots, like $\frac{1}{\sqrt{5}+\sqrt[3]{2}+\sqrt[7]{3}}$? I knew an absolutely awful formula for $3$ cube roots, but aside from that I could only find special cases. Does there exist a general formula? If so, what is it? (I know this question is pointless, but still.)
Apparently it's really difficult to do this by hand for arbitrary radicals. I posted this a long time ago, and since then, I've found a few new formulas through experimentation. $$\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{\left(\sqrt{a}-\sqrt{b}-\sqrt{c}\right)\left(2\sqrt{bc}+a-b-c\right)}{\left(a-b-c\right)^{2}-4bc}$$ $$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}=\frac{\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}-\sqrt[3]{ab}-\sqrt[3]{ac}-\sqrt[3]{bc}\right)\left(\left(3\sqrt[3]{abc}+a+b+c\right)^{2}-3\left(a+b+c\right)\sqrt[3]{abc}\right)}{\left(a+b+c\right)^{3}-27abc}$$ $$\frac{1}{\sqrt[4]{a}+\sqrt[4]{b}+\sqrt[4]{c}}=\frac{\left(\sqrt[4]{a}+\sqrt[4]{b}-\sqrt[4]{c}\right)\left(\sqrt[4]{a}-\sqrt[4]{b}+\sqrt[4]{c}\right)\left(\sqrt[4]{a}-\sqrt[4]{b}-\sqrt[4]{c}\right)\left(2\sqrt{bc}-2\sqrt{ab}-2\sqrt{ac}-a-b-c\right)\left(\left(12a+4b+4c\right)\sqrt{bc}+\left(a-b-c\right)^{2}+4bc\right)}{\left(12a+4b+4c\right)^{2}bc-\left(\left(a-b-c\right)^{2}+4bc\right)^{2}}$$ $$\frac{1}{\sqrt[5]{a}+\sqrt[5]{b}+\sqrt[5]{c}}=\frac{2abc\left(\left(\sqrt[5]{a^{2}}+\sqrt[5]{b^{2}}\right)\sqrt[5]{ab}+\left(\sqrt[5]{a^{2}}+\sqrt[5]{c^{2}}\right)\sqrt[5]{ac}+\left(\sqrt[5]{b^{2}}+\sqrt[5]{c^{2}}\right)\sqrt[5]{bc}-\sqrt[5]{a^{4}}-\sqrt[5]{b^{4}}-\sqrt[5]{c^{4}}-\sqrt[5]{a^{2}b^{2}}-\sqrt[5]{a^{2}c^{2}}-\sqrt[5]{b^{2}c^{2}}-\left(2\sqrt[5]{a}+2\sqrt[5]{b}-3\sqrt[5]{c}\right)\sqrt[5]{abc}\right)\left(\left(\sqrt{4a^{2}b^{2}c^{2}v^{3}+1}-1\right)q_{1}\left(\frac{1}{a^{2}b^{2}c}\right)+2abcq_{2}\left(\frac{1}{a^{2}b^{2}c}\right)\right)p\left(\frac{\sqrt[5]{a^{3}b^{3}c^{4}}}{abc}\right)}{\left(q_{1}\left(\frac{1}{a^{2}b^{2}c}\right)-2abcq_{2}\left(\frac{1}{a^{2}b^{2}c}\right)\right)^{2}-\left(4a^{2}b^{2}c^{2}v^{3}+1\right)q_{1}\left(\frac{1}{a^{2}b^{2}c}\right)^{2}}$$ $$\frac{1}{\sqrt[6]{a}+\sqrt[6]{b}+\sqrt[6]{c}}=\frac{\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}-\sqrt[6]{ab}-\sqrt[6]{ac}-\sqrt[6]{bc}\right)\left(\left(3\sqrt[6]{abc}+\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^{2}-3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt[6]{abc}\right)\left(\left(a+3b+3c\right)\sqrt{a}+\left(3a+b+3c\right)\sqrt{b}-\left(3a+3b+c\right)\sqrt{c}+21\sqrt{abc}\right)\left(2\left(a+3b+3c\right)\left(3a+b+3c\right)\sqrt{ab}+42\left(3a+3b+c\right)\sqrt{abc^{2}}-a\left(a+3b+3c\right)^{2}-b\left(3a+b+3c\right)^{2}+c\left(3a+3b+c\right)^{2}+441abc\right)}{4ab\left(21c\left(3a+3b+c\right)+\left(a+3b+3c\right)\left(3a+b+3c\right)\right)^{2}-\left(a\left(a+3b+3c\right)^{2}+b\left(3a+b+3c\right)^{2}-c\left(3a+3b+c\right)^{2}-441abc\right)^{2}}$$ Where $$p\left(x\right)=\frac{xq\left(x^{5}\right)}{5abcx^{3}-\left(a+b+c\right)x-5}$$ $$q\left(x\right)=u^{5}x^{3}+3Au^{4}x^{2}+3u^{2}\left(A^{2}u+BC\right)x+\left(A^{3}u^{2}-C^{3}u+B^{3}+3ABCu\right)=q_{1}\left(x\right)u+q_{2}\left(x\right)$$ $$A=\left(5v^{4}-5u^{2}v\right)$$ $$B=\left(10u^{2}v^{2}-v^{5}\right)$$ $$C=\left(10uv^{3}-u^{3}\right)$$ $$u=-\frac{\sqrt{4a^{2}b^{2}c^{2}v^{3}+1}+1}{2abc}$$ $$v=-\frac{a+b+c}{15abc}$$ and $q_1(x)$ and $q_2(x)$ have rational coefficients. As for the general case, it seems to involve a matrix inversion. I don't know linear algebra yet, so this wasn't very clear to me. I'm not anywhere near the age I'd need to be to learn it in school, and in fact, most of my peers don't even know how to rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}}$. I really don't feel like waiting for the curriculum to get there, so I have $2$ more questions. Is there a way to do the arbitrary case using only elementary algebra, and where would be a good resource to learn linear algebra in order to do this?
UPDATE: In order to do this in general, use complex numbers so that $n$th roots have $n$ branches. Then take the product over all combinations of the branches. In the example problem, this leads to $2\cdot3\cdot7=42$ possible factors, $1$ of which is the original denominator. For $3$ $n$th roots, the product $$N=\prod_{u,\ v,\ w}^{\ }\left(\sqrt[n]{a_{u}}+\sqrt[n]{b_{v}}+\sqrt[n]{c_{w}}\right)$$ is an integer, where subscripts are used to distinguish branches so it's easier to see what's happening here. In complex exponential form, you have $$N=\prod_{u,\ v,\ w}^{\ }\left(e^{\frac{2\pi ui}{n}}\sqrt[n]{a}+e^{\frac{2\pi vi}{n}}\sqrt[n]{b}+e^{\frac{2\pi wi}{n}}\sqrt[n]{c}\right)$$ This answer was given by Christophe Leuridan.