How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$ How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$


I know $x^3-c^3=(x-c)(x^2+xc+c^2)$ but I can't figure out how to pull this off with $1/3$ instead of $3$
Replace $x$ with $(x^{\frac 13})^3$ and $c = (c^{\frac 13})^3$ and so

$x - c = (x^{\frac 13})^3 - (c^{\frac 13})^3 = (x^{\frac 13} - c^{\frac 13})((x^{\frac 13})^2 + x^{\frac 13}y^{\frac 13} + (c^{\frac 13})^2)$ 

So 

$(x -c) \frac 1{x^{\frac 23} + x^{\frac 13}y^{\frac 13} + c^{\frac 23}}= x^{\frac 13}- c^{\frac 13}$

So $y = \frac 1{x^{\frac 23} + x^{\frac 13}y^{\frac 13} + c^{\frac 23}}$

There IS the assumption that $x^{\frac 23} + x^{\frac 13}y^{\frac 13} + c^{\frac 23} \ne 0$.  But if $x^{\frac 23} + x^{\frac 13}y^{\frac 13} + c^{\frac 23} = 0$ then $x - c = (x^{\frac 13})^3 - (c^{\frac 13})^3 = (x^{\frac 13} - c^{\frac 13})((x^{\frac 13})^2 + x^{\frac 13}y^{\frac 13} + (c^{\frac 13})^2) = 0$ and $x = c$ and $x^{\frac 13} = c^{\frac 13}$ and $y$ could be anything.
Write $w=x-c$.  You want to expand in powers of $w$.  So, as $w \to 0$ we have
\begin{align}
x^{1/3} &amp;= (c+w)^{1/3} = c^{1/3}\left(1+\frac{w}{c}\right)^{1/3}
\\
&amp;= c^{1/3}\left(1+\frac{w}{3c}-\frac{w^2}{9c^2}+\frac{5w^3}{81c^3}+\dots\right)
\\
&amp;= c^{1/3}+\frac{w}{3c^{2/3}}-\frac{w^2}{9c^{5/3}}+\frac{5w^3}{81c^{8/3}}+\dots
\\
x^{1/3} - c^{1/3} &amp;= \frac{w}{3c^{2/3}}-\frac{w^2}{9c^{5/3}}+\frac{5w^3}{81c^{8/3}}+\dots
\\
&amp;= w\left(\frac{1}{3c^{2/3}}-\frac{w}{9c^{5/3}}+\frac{5w^2}{81c^{8/3}}+\dots\right)
\\
&amp;= (x-c)\left(\frac{1}{3c^{2/3}}-\frac{x-c}{9c^{5/3}}+\frac{5(x-c)^2}{81c^{8/3}}+\dots\right)
\end{align}

Of course the last factor is exactly
$$
\frac{x^{1/3}-c^{1/3}}{x-c} ,
$$
which is an analytic function (removable singularity) near $x=c$.