Example 5 What is the minimum degree of the highest term of a polynomial with rational coefficients that has $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots?
(2013, Joint Autonomous Admission Examination of Peking University and Other Universities)
Notice that the polynomial
$$
f(x)=\left(x^{2}-2\right)\left[(x-1)^{3}-2\right]
$$

has roots $\sqrt{2}$ and $1-\sqrt[3]{2}$, and its degree is 5. Therefore, the degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots cannot be less than 5.

If there exists a rational-coefficient polynomial of degree no more than 4
$$
g(x)=a x^{4}+b x^{3}+c x^{2}+d x+e
$$

with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots, where $a, b, c, d, e$ are not all zero, then we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
g(\sqrt{2})=(4 a+2 c+e)+(2 b+d) \sqrt{2}=0, \\
g(1-\sqrt[3]{2}) \\
=-(7 a+b-c-d-e)-(2 a+3 b+2 c+d) \sqrt[3]{2}+ \\
(6 a+3 b+c) \sqrt[3]{4}=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
4 a+2 c+e=0, \\
2 b+d=0, \\
7 a+b-c-d-e=0, \\
2 a+3 b+2 c+d=0, \\
6 a+3 b+c=0 .
\end{array}\right.
\end{array}
$$
(1) + (3) gives
$$
11 a+b+c-d=0.
$$
(6) + (2), (6) + (4) respectively give
$$
\begin{array}{l}
11 a+3 b+c=0, \\
13 a+4 b+3 c=0 .
\end{array}
$$
(7) - (5) gives $a=0$.

Substituting into equations (7), (8) gives $b=c=0$,
Substituting into equations (1), (2) gives $d=e=0$.
Thus, $a=b=c=d=e=0$, which contradicts the assumption that $a, b, c, d, e$ are not all zero.

In conclusion, the minimum degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots is 5.