Elementary method to solve this equation

Question

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.

Answer 1

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 $t_1=1.731...$

Answer 2

If there are any roots, they are all postive, since the function isn't even defined for negative $x$ (it has a $\sqrt x$ in it), and clearly $0$ is not a root.

You can easily see that there is at least one root since $f(0)=\sqrt[3]{24}-\sqrt[3]{6}>0$ but $f(3)=\sqrt[3]{3+\sqrt3}-\sqrt{6}<0$.