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

Question

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 $3$

Answer 1

Yes your way is right, let use that

$$A^3-B^3=(A-B)(A^2+AB+B^2)$$

with $A^3=x$ and $B^3=c$.

Then

$$x^{1/3}-c^{1/3}=(x-c)\cdot \frac1{x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}}$$

Answer 2

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.

Answer 3

Write $w=x-c$. You want to expand in powers of $w$. So, as $w \to 0$ we have \begin{align} x^{1/3} &= (c+w)^{1/3} = c^{1/3}\left(1+\frac{w}{c}\right)^{1/3} \\ &= c^{1/3}\left(1+\frac{w}{3c}-\frac{w^2}{9c^2}+\frac{5w^3}{81c^3}+\dots\right) \\ &= 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} &= \frac{w}{3c^{2/3}}-\frac{w^2}{9c^{5/3}}+\frac{5w^3}{81c^{8/3}}+\dots \\ &= w\left(\frac{1}{3c^{2/3}}-\frac{w}{9c^{5/3}}+\frac{5w^2}{81c^{8/3}}+\dots\right) \\ &= (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 $x=c$.