Solving : $2n^{\frac{1}{2}}-1.5n^{\frac{1}{3}}+n^{\frac{1}{6}}=y$ for $n$ in terms of $y$ I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{\frac{1}{2}}-bn^{\frac{1}{3}}+cn^{\frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no avail. 
Here is what I have tried so far in solving the following equation for $n$ in terms of $y$:
$2n^{\frac{1}{2}}-1.5n^{\frac{1}{3}}+n^{\frac{1}{6}}=y$
I first tried to rewrite it as a quadratic but after pulling out $n^{\frac{1}{6}}$ like this:
$n^{\frac{1}{6}}(2n^{\frac{1}{3}}-1.5n^{\frac{1}{6}}+1)=y$ 
But then there is nothing further I can do to factor this.
I couldn't find any power that I could raise $n$ to multiply both sides either.
Any advice as how to proceed in solving this would be greatly appreciated.
Thank you.
 A: Let $x=n^{1/6}$ (i.e. $n=x^6$):
$$2x^3-\frac 3 2x^2+x=y$$
Subtract $y$ and divide both sides by $2$:
$$x^3-\frac 3 4x^2+\frac 1 2x-\frac 1 2y=0$$
In order to reduce this to a depressed cubic, let $x=z+\frac 1 4$:
$$z^3+\frac{5z}{16}+\frac{3}{32}-\frac 1 2y=0$$
In order to turn this into a quadratic, let $z=w+\frac{5}{48w}$:
$$w^3+\frac{3}{32}-\frac 1 2y-\frac{125}{110592w^3}=0$$
Multiply by $w^3$:
$$w^6+\left(\frac{3}{32}-\frac 1 2y\right)w^3-\frac{125}{110592}=0$$
Time to apply the quadratic formula:
$$w^3=\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}$$
Now, there are three complex solutions to this equation. To represent this, I will use $\omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $\omega=1$, one for $\omega=\frac{-1+\sqrt{-3}}{2}$, and one for $\omega=\frac{-1-\sqrt{-3}}{2}$:
$$w=\omega\sqrt[3]{\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}}$$
From here, I leave the rest to you: Solve for $z$ using $z=w-\frac{5}{48w}$, solve for $x$ using $x=z+\frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!
A: Using $x=n^6$ as Noble Mushtak did and follow the steps given here, using the hyperbolic solution for one real root,  you should arrive to the simple equation
$$\color{blue}{x=\frac{1}{4}-\frac{1}{2} \sqrt{\frac{5}{3}} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{1}{192} \sqrt{\frac{5}{3}} \left(\frac{81}{4}-108
   y\right)\right)\right)}$$ which is positive if $y >0$.
