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

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 $\sqrt[5]{1782+405t}-t=2.$
