How can I find a polynomial with $\sqrt{2}+\sqrt[3]{3}+\sqrt[5]{5}$ as a root, analytically? I know that the degree will be $30.$
My question is unique because the root of the polynomial is the sum of three distinct roots of three distinct numbers.
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possible duplicate of Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ – MJD Jun 12 '15 at 05:57
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It should be noted that the "duplicate" question does not ask the same question as this one, but the accepted answer does use a method that could also answer this question. – Rory Daulton Jun 12 '15 at 10:50
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Let $a_1$, $a_2$ be the two square roots of $2$, $b_1$, $b_2$, $b_3$ the three cubir roots of $3$ and $c_1$, $\dots$, $c_5$ the five fifth roots of $5$. The polynomial $$\prod_{\substack{1\leq i\leq 2\\1\leq j\leq 3\\1\leq k\leq 5}}(x-(a_i+b_j+c_k))$$ has your element as root. Its coefficients are symmetric functions of the $a$s, of the $b$s and of the $c$s (separately) so they can be expressed in term of the coefficients of the polynomials $X^2-2$, $X^3-3$ and $X^5-5$.
Mariano Suárez-Álvarez
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Another way to do this: let $A$, $B$, $C$ be the companion matrices of $x^2-2$, $x^3-3$ and $x^5 - 5$, let $I_n$ be the $n \times n$ identity matrix, let $M = A \otimes I_3 \otimes I_5 + I_2 \otimes B \otimes I_5 + I_2 \otimes I_3 \otimes C$ (a $30 \times 30$ matrix). The characteristic polynomial of $M$ is what you want. Explicitly, according to Maple, $$ {x}^{30}-30\,{x}^{28}-30\,{x}^{27}+420\,{x}^{26}+510\,{x}^{25}-3235\,{ x}^{24}-3600\,{x}^{23}+16350\,{x}^{22}+8040\,{x}^{21}-218301\,{x}^{20} +37650\,{x}^{19}+1609510\,{x}^{18}-1147230\,{x}^{17}-2931060\,{x}^{16} -12570136\,{x}^{15}-2033490\,{x}^{14}+82639140\,{x}^{13}+19899460\,{x} ^{12}-161777130\,{x}^{11}-191993163\,{x}^{10}+135990670\,{x}^{9}+ 169540905\,{x}^{8}-206696910\,{x}^{7}+648301895\,{x}^{6}+1291380294\,{ x}^{5}-362679480\,{x}^{4}-3167370380\,{x}^{3}-115218555\,{x}^{2}+ 1105068720\,x+867818606 $$
Robert Israel
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