# 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.

• Hint: Try a substitution like $x = n^{\frac{1}{6}}$. – John Omielan Dec 30 '18 at 2:44

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 $$\omega=\frac{-1-\sqrt{-3}}{2}$$:

$$w=\omega\sqrt{\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}}$$

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!

• Thank you for this response, it worked well! – limitsandlogs224 Dec 30 '18 at 16:49

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 $$y >0$$.