<s>[INST] How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers. Prove that
$$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}\in N$$
This problem from this
My try: let 
$$x=\sqrt[3]{35+15\sqrt{6}}+\sqrt[3]{35-15\sqrt{6}}$$
then 
$$x^3=70+3\sqrt[3]{35^2-(15\sqrt{6})^2}\cdot x=70-3\sqrt[3]{125}x$$
then we have
$$(x^3-70)^3=-27\cdot 125 x$$
$$\Longrightarrow x^6-15x^4-140x^3+225x^2+1050x+4900=0$$
Then I can't ,and I think this is interesting problem, Thank you
 [/INST] 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.
</s>