Create a monic polynomial with integer coefficients with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a root. Create a monic polynomial with integer coefficients with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a  root.
I want to create such polynomial but all the ways I have tried don't give a polynomial with integer coefficients and all of them give me irrational or non integer rationals. How should I work?
Please use high school math for solution and don't use matrix.
 A: The systematic approach would be to write $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a system of algebraic equations:
$$
\begin{align}
\begin{cases}
x - a - b - c & = 0 \\
a^3 - 2 & = 0 \\
b^3 - 3 & = 0 \\
c^3 - 5 & = 0
\end{cases}
\end{align}
$$
Eliminating $a,b,c$ between the equations will give a single equation in $x$ with integer coefficients.
Elimination can be done by repeatedly using the resultant of two polynomials, which is a routine calculation, thogh not pretty to do by hand in this particular case. Wolfram Alpha resultant[ resultant[ resultant[ x - a - b - c, a^3 - 2, a], b^3 - 3, b ], c^3 - 5, c] gives:
$$-x^{27} + 90 x^{24} - 1089 x^{21} + 62130 x^{18} - 105507 x^{15} + 16537410 x^{12} + 30081453 x^{9} + 1886601330 x^{6} - 73062900 x^{3} + 6859000$$
A: Hint: Suppose you want $\sqrt{2}+\sqrt3$ as the root of a polynomial with integer coefficients. Then we have:
\begin{align}
x&=\sqrt2+\sqrt3\\
x^2&=5+2\sqrt6\\
x^2-5&=2\sqrt6\\
(x^2-5)^2&=24\\
(x^2-5)^2-24&=0.
\end{align}
Thus, $\sqrt2+\sqrt3$ is a root of $$f(x)=x^4-10x^2+1.$$
A: Let $u=\cos(2\pi/3)+i\sin(2\pi/3)$. This is a third primitive root of unity, $u^3=1$. Next, let $a=\sqrt[3]2$, $b=\sqrt[3]3$, and $c=\sqrt[3]5$. The desired polynomial can be taken in the form
$$
 P(x)\equiv P(x,a,b,c)=\prod_{j,k,l=1}^3(x-u^ja-u^kb-u^lc) 
$$
and has degree $27$ (because there are $3\times3\times3=27$ factors in the product). Indeed,
$\bullet$ It is clearly monic.
$\bullet$ If we replace $a$ by $ua$ in the definition of the polynomial, then it does not change (the factors are just permuted, $a\to ua$, $ua\to u^2a$, $u^2a\to u^3a=a$). On the other hand, the coefficient of $a^s$ in $P(x,a,b,c)$ is multiplied by $u^s$. Thus, if $s$ is not a multiple of $3$, then the coefficient of $a^s$ is zero. In other words, only $a^3=2$, $a^6=4$ etc. occur in the coefficients. The same is true of $b$ and $c$. Thus, the coefficients of $P(x)$ are sums of products of integers by powers of $u$. 
$\bullet$ The coefficients of the polynomial $P(x)$ are real (because complex conjugation just interchanges $u$ and $u^2$). Let $R$ be any of the coefficients. By the preceding, $R=m+nu+pu^2$, where $m,n,$ and $p$ are integers; since $R$ is real, we have $p=n$, and $R=m+n(u+u^2)=m-n$ is an integer. Thus, the polynomial has integer coefficients. 
$\bullet$ Clearly, $a+b+c$ is a root of $P(x)$.
Thus, $P(x)$ satisfies all the necessary requirements. 
