Number of real positive roots of a polynomial? Consider the polynomial
$$f(x)=x((1+x^n)^n+a^n)-a(1+x^n)^n,$$
where $n\geq 2$ is a positive integer and $a$ is a positive real number. I'm interesting in deducing the number of positive real roots of $f$ as a function of $a$ and $n$. Checking numerically, it seems that if $a$ is sufficiently small in comparison to $n$ there is a single positive root, otherwise there are three.
Initially, I thought of simply using Descartes' rule of signs. Applying the binomial theorem and re-arranging we have that
$$f(x)=\begin{pmatrix}n\\0\end{pmatrix}x^{n^2+1}-a\begin{pmatrix}n\\0\end{pmatrix}x^{n^2}+\begin{pmatrix}n\\1\end{pmatrix}x^{n^2-n+1}-a\begin{pmatrix}n\\1\end{pmatrix}x^{n^2-n}+\dots+\left[\begin{pmatrix}n\\n\end{pmatrix}+a^n\right]x-a\begin{pmatrix}n\\n\end{pmatrix}$$
from which it is straightforward to see that $f$ has $2n+1$ sign changes, that is, the number of positive real roots of $f$ is $2n+1$ or less than $2n+1$ by an even number.
Are there any other relevant results that could be used to narrow down the number of positive roots any further (with the aim of proving or disproving that there are either $1$ or $3$)? 

I've stumbled onto the value of $a$ at which $f(x)$ transitions from one positive real zero to three positive real zeros.  I have no proof though.

When
  $$
a = \frac{n}{(n-1)^{1+1/n}},
$$
  $f$ has a triple zero at
  $$
x = \frac{1}{(n-1)^{1/n}}.
$$

So suppose that $a$ has this value.  We will show that
$$
f\left(\frac{y}{(n-1)^{1/n}}\right) = (n-1)^{-\frac{1+n+n^2}{n}} \left[n^n y+\Bigl((n-1)y-n\Bigr) \left(n-1+y^n\right)^n\right]
$$
has a triple zero at $y=1$.  For convenience let
$$
g(y) = n^n y+\Bigl((n-1)y-n\Bigr) \left(n-1+y^n\right)^n.
$$
We can check that $g$ has a zero at $y=1$ by simply substituting it in.  Differentiating yields
$$
g'(y) = n^n + (n-1)\left(n-1+y^n\right)^n + n^2 y^{n-1} \Bigl((n-1)y-n\Bigr) \left(n-1+y^n\right)^{n-1},
$$
and we can again check that this has a zero at $y=1$.  Differentiating once more yields
$$
\begin{align}
g''(y) &= n^2 (n-1) y^{n-2} (n-1+y^n)^{n-1} \\
&\qquad \times \Bigl(n-n^2+(n^2-1)y-(n^2+n)y^n+(n^2+1)y^{n+1}\Bigr),
\end{align}
$$
which also has a zero at $y=1$.
These values for $a$ and $x$ were found numerically with the help of the Inerse Symbolic Calculator.  I don't see how to use this information to answer the problem in its entirety.  Perhaps someone else will have an idea.
