About a polynomial with a root $\sqrt[3]{2} + \sqrt{5}$.

Question

How to find a polynomial with integer coefficients and a root $\sqrt[3]{2} + \sqrt{5}$? The degree of this polynomial can't be more than 7.

Answer 1

Just write $x=\sqrt[3]{2} + \sqrt{5}$, and start computing $x-\sqrt{5}=\sqrt[3]{2}$. Now cube this equation. Repeating this procedure finally you will have a polynomial with integer coefficients. It certainly has the required root, but one has to check that it is irreducible.

Answer 2

Let $$x=\sqrt[3] 2 + \sqrt 5 $$

For getting integer coefficients, you can do the following operation : $$(x-\sqrt 5 )^3=2 $$

Now, collect all the terms under the square root in LHS, and radical free terms on RHS, and square both the sides. You'll get all the coefficients as integers, and the degree will be $6$.

Answer 3

Consider the column vector $$v=\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4\\\sqrt5\\\sqrt[3]2\sqrt5\\\sqrt[3]4\sqrt5}.$$ Then $Av=\sqrt[3]2v$ and $Bv=\sqrt5v$ where $$A=\pmatrix{0&1&0&0&1&0\\0&0&1&0&0&1\\2&0&0&2&0&0\\ 0&1&0&0&1&0\\0&0&1&0&0&1\\2&0&0&2&0&0}$$ and $$B=\pmatrix{0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\ 5&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0}.$$ Then $Cv=(\sqrt[3]2+\sqrt5)v$ where $C=A+B$, so that $\sqrt[3]2+\sqrt5$ is an eigenvalue of $C$. Take the characteristic polynomial of $C$.

Answer 4

Let $P(x)=x^3-2$ and $Q(x)=x^2-5$ then $\sqrt[3]{2} + \sqrt{5}$ is a root of the resultant of $P(x)$ and $Q(t-x)$ which is an integer polynomial in $t$: $$\text{resultant}(P(x),Q(t-x))=t^6-15t^4-4t^3+75t^2-60t-121.$$