Elementary method to solve this equation I have to describe ''the sign of the root(s)''$$\sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-x}-\sqrt[3]{6}=0$$ for k-11 students , who did not learned derivation. 
I can solve the equation by taking $f(x)=\sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-x}-\sqrt[3]{6}$  also by graphing https://www.desmos.com/calculator/u0pbqp1hvu
but , is there a simple trick to show the sign of roots(s) .
Remark:by derivation or graphing we can see 1 positive root exists . 
thanks in advance.

Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and since
$$3+\sqrt{x}=3-x=-6$$ is impossible,
our equation is equivalent to
$$3+\sqrt{x}+3-x-6+3\sqrt[3]{6(3+\sqrt{x})(3-x)}=0$$ or
$$(\sqrt{x}-x)^3+162(3+\sqrt{x})(3-x)=0,$$
which after substitution $\sqrt{x}=t$ gives
$$t^6-3t^5+3t^4+161t^3+486t^2-486t-1458=0$$ and the last equation has unique non-negative root $t_1=1.731...$
