Finding a polynomial with integer/rational coefficients and a given algebraic root

Question

Just for fun, I have been attempting to find polynomials with integer coefficients from a given root. I have been able to do this with square roots: for example, if say I wanted to find a polynomial with a root $\sqrt 2$, we would set an equation like this: $$x-\sqrt 2=0$$ Then just move the square root to the other side, square, and end with a result of $$x=\sqrt 2$$ $$x^2=2$$ $$x^2-2=0$$ This strategy can also be done if given a single nth root, or multiple square roots, or a single nth root along with any number of square roots. However, I'm not quite sure how to tackle multiple nth roots like say $3^\frac{1}{3} + 5^\frac{1}{3}$ or $3^\frac{1}{3} + 5^\frac{1}{5}$. Just taking the first example, if we start with the equation $$x-3^\frac{1}{3}-5^\frac{1}{3}=0$$ and move the cube root of 5 to the other side and cube it, we get $$x-3*3^\frac{1}{3}x^2+3*3^\frac{2}{3}x-3=5$$ After getting here I'm unsure of where to go next, as I don't know how to handle both the $3^\frac{1}{3}$ and $3^\frac{2}{3}$ at the same time. I would really appreciate it if someone could help with both this problem and a more general problem solving technique for ones like the other example I showed or something more complicated like say $\frac{3^\frac{2}{3}+5^\frac{1}{5}}{7^\frac{1}{7}+\sqrt 2}$

Answer 1

Let $\,x=\sqrt[3]{3}+\sqrt[3]{5}\,$, then using the identity $\,(a+b)^3=a^3+b^3+3ab(a+b)\,$:

$$ x^3 = \left(\sqrt[3]{3}\right)^3+\left(\sqrt[3]{5}\right)^3 + 3 \cdot \sqrt[3]{3}\sqrt[3]{5}\cdot(\sqrt[3]{3}+\sqrt[3]{5}) = 3+5+3 \cdot \sqrt[3]{15} \cdot x = 8 + 3 \sqrt[3]{15} \, x $$

It follows that $\,x^3-8=3 \sqrt[3]{15} \, x\,$, and therefore:

$$(x^3-8)^3=3^3 \cdot 15 \cdot x^3 \quad\iff\quad x^9 - 24 x^6 - 213 x^3 - 512 = 0$$

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[ EDIT ] Radicals are algebraic numbers, and the algebraic numbers form a field (i.e. closed under addition/multiplication and their inverses), so any rational expression in algebraic numbers is itself algebraic. A polynomial with integer coefficients having such an expression as a root (though not necessarily the minimal one) can be determined algorithmically using polynomial resultants.

For example, one would write $\,x = \displaystyle\frac{\sqrt[3]{9}+\sqrt[5]{5}}{\sqrt[7]{7}+\sqrt 2} = \frac{a+b}{c+d}\,$ as the polynomial system:

$$ \begin{cases} \begin{align} cx+dx-a-b=0 \\ a^3-9=0 \\ b^5-5=0 \\ c^7-7=0 \\ d^2 - 2 = 0 \end{align} \end{cases} $$

Then, using resultants repeatedly, the variables $\,a,b,c,d\,$ can be successively eliminated from the system, leaving in the end an equation in $\,x\,$ with integer coefficients. (However, the calculations are not pretty, and would normally be done using some CAS rather than by hand.)

Answer 2

What follows is a "school algebra" way to rationalize $\;r_1\sqrt[3]{s_1} + r_2\sqrt[3]{s_2} + r_3\sqrt[3]{s_3},\;$ where $r_1,$ $r_2,$ $r_3,$ $s_1,$ $s_2,$ and $s_3$ are rational numbers. Incidentally, I am not assuming that any of $\sqrt[3]{s_1},$ $\sqrt[3]{s_2},$ or $\sqrt[3]{s_3}$ is irrational.

Note that it suffices to rationalize $\;\sqrt[3]{k} + \sqrt[3]{m} + \sqrt[3]{n},\;$ where $k,$ $m,$ and $n$ are rational numbers (e.g. write $\;r_1\sqrt[3]{s_1} = \sqrt[3]{(r_1)^3s_1},\;$ etc.). In fact, without loss of generality, we can assume $k,$ $m,$ and $n$ are integers, although doing so does not make any difference in what follows. Actually, the reduction to $\sqrt[3]{k} + \sqrt[3]{m} + \sqrt[3]{n}$ is not needed. I only did it to reduce the symbol clutter.

Let $\;a = \sqrt[3]{k}\;$ and $\;b = \sqrt[3]{m}\;$ and $\;c = \sqrt[3]{n}.\;$ Then

$$ (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$

$$ = \;\; (k + m + n) - 3\sqrt[3]{kmn} \;\; = \;\; (k+m+n) -\sqrt[3]{27kmn} $$

Let $\;x = (k+m+n)\;$ and $\;y = \sqrt[3]{27kmn}.\;$ Then

$$ (x-y)(x^2 + xy + y^2) \;\; = \;\; x^3 - y^3 \;\; = \;\; (k + m + n)^3 - 27kmn $$

Therefore,

$$ \frac{1}{\sqrt[3]{k} + \sqrt[3]{m} + \sqrt[3]{n}}$$

can be rationalized by multiplying both the numerator and the denominator by

$$ (a^2 + b^2 + c^2 - ab - bc - ac)(x^2 + xy + y^2)$$

where $a,$ $b,$ $c,$ $x,$ and $y$ are expressions involving $k,$ $m,$ and $n,$ as indicated above. The result will be

$$ \frac{(a^2 + b^2 + c^2 - ab - bc - ac)(x^2 + xy + y^2)}{(k + m + n)^3 - 27kmn} $$