algebra precalculus - Complicate the expression for $frac{p-1}{p}x+frac{a}{p}x^{1-p}-a^{1/p}$

I am interested in finding a way to go between the RHS and LHS of the following equation
$$\frac{p-1}{p}x+\frac{a}{p}x^{1-p}-a^{1/p}=(x-a^{1/p})\left[\frac{p-1}{p}-\frac{1}{p}\left(\sum_{n=1}^{p-1}\left(\frac{a^{1/p}}{x}\right)^n\right)\right]$$
In the book, it is said it is an "easily" derived formula, however, even after I reversed engineeres the RHS (recognizing geometric sum and simplifying as much as possible) I have been unable to find a way from the left to the right side.

I feel like I must be missing something since this manipulation was supposed to be easy but I do not see it. How would I have gone from the LHS to the RHS?

Answer

Working on another way round usually is not easy.

Observe that

\begin{array}{rcl}
X^2-2X+1 &=& (X-1)(X-1) \\
2X^3-3X+1 &=& (X-1)(2X^2-X-1) \\
3X^4-4X+1 &=& (X-1)(3X^3-X^2-X-1) \\
4X^5-5X+1 &=& (X-1)(4X^4-X^3-X^2-X-1) \\
& \vdots & \\
(p-1)X^{p}-pX^{p-1}+1 &=& (X-1)[(p-1)X^{p-1}-X^{p-2}-\ldots-X-1] \\
X &=& \dfrac{x}{a^{1/p}} \\
\dfrac{p-1}{p}x+\dfrac{a}{p}x^{1-p}-a^{1/p} &=&

\dfrac{(p-1)X^{p}-pX^{p-1}+1}{p}\cdot \dfrac{a^{1/p}}{X^{p-1}} \\
&=& a^{1/p}(X-1)
\left[
\dfrac{p-1}{p}-\dfrac{1}{p}
\left( 1+\dfrac{1}{X}+\ldots+\dfrac{1}{X^{p-1}} \right)
\right]
\end{array}

The result follows.

Further point to be noticed:
\begin{align*}
nx^{n+1}-(n+1)x^{n}+1 &= (x-1)(nx^{n}-x^{n-1}-\ldots-x-1) \\
&= (x-1)^2[nx^{n-2}+(n-1)x^{n-3}+\ldots+2x+1]
\end{align*}