อ้างอิง:
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Problem. Determine a polynomial $P(x)$ with integer coefficients such that $P(x)$ is of the lowest degree and $P(x)$ has $\sqrt{2}+\sqrt[3]{3}$ as a root.
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This answer is a continuation of the previous answer by ครูนะ above. While $$P(x):=x^6-6x^4-6x^3+12x^2-36x+1$$ is indeed an irreducible element of $\mathbb{Z}[x]$, it is not sufficient to prove its irreducibility by merely inspecting the rational roots. This polynomial may very well factor into a product of three irreducible quadratic polynomials, or two irreducible cubic polynomials. Here is an approach which shows that $P(x)$ is indeed irreducible (over $\mathbb{Q}$ or $\mathbb{Z}$).
Suppose that $P(x)$ is reducible. We know that it has no rational roots, whence $P(x)$ has no linear factor over $\mathbb{Q}$. Reducing $P(x)$ modulo $3$ yields $$P(x)\equiv x^6+1\equiv (x^2+1)^3\pmod{3}\,.$$ Therefore, if $P(x)$ is reducible, then it has an irreducible quadratic factor $Q(x)\in\mathbb{Z}[x]$. We may assume that $Q(x)$ is monic. Thus, $$Q(x)\equiv x^2+1\pmod{3}\,,$$ so it follows that
$$Q(x)=x^2+ax+1$$
for some integer $a$ divisible by $3$. If you divide $P(x)$ by $Q(x)$, then the remainder is
$$-(a^5-10a^3+6a^2+27a+30)\,x-(a^4-9a^2+6a+18)\,.$$
Since $Q(x)$ is a factor of $P(x)$, the remainder must be $0$. Therefore,
$$a^5-10a^3+6a^2+27a+30=0$$
and
$$a^4-9a^2+6a+18=0\,.$$
Therefore, the polynomial
$$f(x):=x^5-10x^3+6x^2+27x+30$$
and
$$g(x):=x^4-9x^2+6x+18=(x+3)(x^3-3x^2+6)$$
have a common factor $x-a$. However, $a\neq -3$ since $f(-3)=30\neq 0$. Therefore, $x-a$ must be a factor of $x^3-3x^2+6$. Nonetheless, $x^3-3x^2+6$ has no rational root, which can be proven via the Rational Root Theorem.
Therefore, $P(x)$ is irreducible. Thus, it is a polynomial of the lowest degree with $\sqrt{2}+\sqrt[3]{3}$ as a root. Indeed, all roots of $P(x)$ are of the form
$$s\sqrt{2}+t\sqrt[3]{3}\,,$$
where $s\in\{-1,+1\}$ and $t\in\left\{1,\dfrac{-1+\sqrt{-3}}{2},\dfrac{-1-\sqrt{-3}}{2}\right\}$.