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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.

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

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  • $\begingroup$ Thank you for this response, it worked well! $\endgroup$ – limitsandlogs224 Dec 30 '18 at 16:49
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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 $y >0$.

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