I have this root and must find a polynomial with all integer coefficients that has this as a root: $$\sqrt[3]{2 + \sqrt{3}} + \sqrt5$$
I understand that the first step is to set this equal to x and eliminate roots one by one, however for this particular root it gets messy following this simple step. I was wondering what the best way to solve this problem is. Any help is appreciated!!
In Maple:
evala(Norm(z-((2+sqrt(3))^(1/3) + sqrt(5)));
Sort of by hand: Let $ t = (2 + \sqrt{3})^{1/3} + \sqrt{5}$.
Let's start with $x_1 = 2 + \sqrt{3}$. This satisfies $(x_1 - 2)^2 = 3$.
Next $x_2 = x_1^{1/3}$. So $x_1 = x_2^3$, i.e. $(x_2^3 - 2)^2 = 3$, or $ x_2^6 - 4 x_2^3 + 1 = 0$.
Now $x_3 = \sqrt{5}$, satisfying $x_3^2 = 5$.
The tricky part is $t = x_2 + x_3$. $x_2$ is an eigenvalue of the companion matrix to the polynomial $x^6 - 4 x^3 + 1$, $$ A_2 = \pmatrix{0 & 0 & 0 & 0 & 0 & -1 \cr 1 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr 0 & 0 & 1 & 0 & 0 & 4 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 0 & 0 & 0 & 0 & 1 & 0 }$$
$x_3$ is an eigenvalue of the companion matrix to its polynomial $x^2-5$: $$ A_3 = \pmatrix{0 & 5\cr 1 & 0\cr} $$ So $x_2 + x_3$ is an eigenvalue of $$ A_2 \otimes I_2 + I_6 \otimes A_3 = \pmatrix{0 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \cr 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \cr 1 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 1 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 1 & 0 & 0 & 5 & 0 & 0 & 4 & 0 \cr 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 4 \cr 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 5 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 5 \cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 }$$ The characteristic polynomial of this matrix is your answer, but I'd hate to do this by hand: $$z^{12}-30 z^{10}-8 z^{9}+375 z^{8}-2482 z^{6}+1200 z^{5}+9285 z^{4}-8008 z^{3}-16800 z^{2}+14880 z +13876$$