How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers.

Denotes $t=\sqrt[3]{15 \sqrt{6}+35}+\sqrt[3]{35-15 \sqrt{6}},$ then $t^3+15 t-70=0.$

$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}$

$=\sqrt[5]{1782+405t}-t$

Since $(t+2)^5-(1782+405t)=(t+5)^2 \left(t^3+15 t-70\right)=0$, we get $\sqrt[5]{1782+405t}-t=2.$

This isn't a complete answer, but I think it might be enough for you to go on:

Note that $x=-3$ is a root of the polynomial. Therefore $x+3$ divides $x^5-405x-972$. Dividing yields $x^4-3x^3+9x^2-27x-324$. It turns out that $x=-3$ is a root of this quotient as well. Dividing again by $x+3$ yields $x^3-6x^2+27x-108$. The exact roots of the cubic can be found, and $\sqrt[3]{35\pm\sqrt{6}}$ appears in those roots.