Child Labor 13

Substitute $x=y^{12}$ to get rid of those pesky radicals.Then the equation becomes: $\begin{aligned} y^{12}=y^6+y^4+y^3\\ y^{12}-y^6-y^4-y^3=0\\ \end{aligned}$ Factoring the equation,we get $y^3(y+1)(y^8-y^7+y^6-y^5+y^4-y^3-1)$ .The triple root $y=0$ is discarded since that would give $x=0^{12}=0$ but we are given that $x>0$ .The root $y=-1$ is also discarded since $x=(-1)^{12}=1$ does not satisfy the equation.Solving $y^8-y^7+y^6-y^5+y^4-y^3-1$ by any root-finding algorithm gives two real roots: $y=1.1566$ or $y=-0.698644$ . $x=(-0.698644)^{12}$ does not satisfy the equation,hence the answer is $x=(1.1566)^{12}\approx \boxed{5.73057}$ to $5$ decimal places.

Graphing and using zoom box, there are two solutions, 0 and 5.7305778. I used ti -83 plus.