Solving the trinomial Consider the following trinomial equation
\begin{equation}
ax^n-x+1=0.
\end{equation}
If $a = \left(\frac{b^{n-1}-1}{b^n-1}\right)^{n-1}\left(1-\frac{b^{n-1}-1}{b^n-1}\right)$ for some $b\in\mathbb{R}$, then one can check that a solution to the above equation is
\begin{equation}
x = \frac{b^n-1}{b(b^{n-1}-1)}.
\end{equation}
Although this is a valid solution, does this actually bring any original insight to the solution of the trinomial above? Since to solve it for some arbitrary $a$ we need to invert the function $f(b) = \left(\frac{b^{n-1}-1}{b^n-1}\right)^{n-1}\left(1-\frac{b^{n-1}-1}{b^n-1}\right)$. 
 A: As you observed, this more or less replaces solving one equation by a possibly more complicated one. It won't help in finding an algebraic formula (= one involving root extractions and basic arithmetic) for the zeros, for the theory saying that such a formula cannot exist is immune to tricks like this.
Do observe that the general theory explaining that no formula exists for the zeros of equations of degrees five or higher is about generic equations. Of course, we can cook up special cases where the roots can be extracted – just plug in some nice value of $b$ into your formula to get a special value of $a$!
But, even a simple choice of $a$ does not help. This recent thread shows that the choice $a=1$ leads to a polynomial whose roots cannot be written in terms of radicals when $n\ge5$.
Understanding that thread does require basic Galois theory (typically third to fifth year college material).

Your observation is more about rationally parametrizing the curve $C$ defined by the equation
$$
yx^n-x+1=0.\qquad (*)
$$
A parametrization $(x,y)=(x(b),y(b))$ being given by
$$
\begin{aligned}
y(b)&=\left(\frac{b^{n-1}-1}{b^n-1}\right)^{n-1}\left(1-\frac{b^{n-1}-1}{b^n-1}\right),\\
x(b)&= \frac{b^n-1}{b(b^{n-1}-1)}.
\end{aligned}
$$
The existence of such a rational parametrization is equivalent to $C$ having genus $g=0$.
Whenever we can solve one parameter as a rational function of the other (as is the case with $C$) this is automatic. There are less obviously parametrizable curves. The most frequently occurring is the unit circle 
$$x^2+y^2=1,$$
which has the famous parametrization
$$
\begin{aligned}
x(t)&=\frac{2t}{1+t^2},\\
y(t)&=\frac{1-t^2}{1+t^2}.
\end{aligned}
$$
This parametrization leads to those nice substitutions for calculating a number of integrals as well as to the parametrization of Pythgorean triples.
Some singular genus zero cubics also appear frequently in calculus textbooks.
The $\alpha$-curve
$$
y^2=x^3+x^2
$$
has a parametrization $x(t)=t^2-1$, $y(t)=t(t^2-1)$, and the Folium
$$
x^3+y^3=3xy
$$
has a parametrization
$x(t)=3t/(1+t^3), y(t)=3t^2/(1+t^3).$

But, the simplest way to rationally parametrize $C$ is surely to use $x$ as a parameter, and solve
$$
y=\frac{x-1}{x^n}
$$
from $(*)$.
Your parametrization is gotten from this by plugging in
$x=x(b)$.
