If $x=10+6\sqrt{3}$, then $\sqrt[3]{x}-1$ is

$\begin{array}{l}x=10+6\sqrt{3}\\ \sqrt[3]{x}=\sqrt[3]{10+6\sqrt{3}}=a+\sqrt{b}\\ \sqrt[3]{10+6\sqrt{3}}=a+\sqrt{b}\end{array}$

Cubing on both sides

$\begin{array}{l}{\left(\sqrt[3]{10+6\sqrt{3}}\right)}^3={(a+\sqrt{b})}^3\\ 10+6\sqrt{3}=a^3+3ab+3a^2\sqrt{b}+{\left(\sqrt{b}\right)}^3\\ a^3+3ab=10\cdots \left(1\right)\\ \sqrt[3]{10+6\sqrt{3}}=a+\sqrt{b},\sqrt[3]{10-6\sqrt{3}}=a-\sqrt{b}\\ a^2-b=\sqrt[3]{100-108}\\ b=a^2+2\cdots \left(2\right)\end{array}$

$\begin{array}{l}\text{by\hspace{0.17em}\hspace{0.17em}}\left(1\right)\left(2\right),a^3+3a(a^2+2)=10\Rightarrow 4a^3+6a=10\\ a=1,b=3\\ \therefore \sqrt[3]{x}=1+\sqrt{3}\end{array}$

$\sqrt[3]{x}-1=\sqrt{3}=$ Irrational