Comment upon nature of the roots How many roots are there of the following polynomial? How many are real, and how many are complex?
$$\left(x-3\right)^9+\left(x-3^2\right)^9+\left(x-3^3\right)^9+\left(x-3^4\right)^9+...+\left(x-3^9\right)^9=0$$

I started by trying a simpler version of the same problem to see if the answer shows a pattern which can be used to solve this, more difficult version:
$$(x-3)^3 + (x-3^2)^3 + (x-3^3)^3 = 0$$
Then expanding the brackets. To be honest, the 3 just started confusing matters while looking for a pattern so I've replaced it with a y (I'll substitute it back later)
$$\begin{align}(x-y)^3 &= (x-y)(x-y)(x-y)\\
&= x(x-y)(x-y) - y(x-y)(x-y)\\
&= x^2(x-y) - xy(x-y) - yx(x-y) + y^2(x-y)\\
&= x^3 - x^2y - x^2y + xy^2 - x^2y + xy^2 + xy^2 - y^3\\
&= x^3 - 3x^2y + 3xy^2 - y^3\end{align}$$
$$\begin{align}(x-y^2)^3 &= (x-y^2)(x-y^2)(x-y^2)\\
&= x(x-y^2)(x-y^2) - y^2(x-y^2)(x-y^2)\\
&= x^2(x-y^2) - xy^2(x-y^2) - xy^2(x-y^2) + y^4(x-y^2)\\
&= x^3 - x^2y^2 - x^2y^2 + xy^4 - x^2y^2 + xy^4 + xy^4 - y^6\\
&= x^3 - 3x^2y^2 + 3xy^4 - y^6\end{align}$$
$$\begin{align}(x-3^2)^3 &= (x-y^3)(x-y^3)(x-y^3)\\
&= x(x-y^3)(x-y^3) - y^3(x-y^3)(x-y^3)\\
&= x^2(x-y^3) - xy^3(x-y^3) - xy^3(x-y^3) + y^6(x-y^3)\\
&= x^3 - x^2y^3 - x^2y^3 + xy^6 - x^2y^3 + xy^6 + xy^6 - y^9\\
&= x^3 - 3x^2y^3 + 3xy^6 - y^9\end{align}$$
And I get... 
$$x^3 - 3x^2y + 3xy^2 - y^3 +x^3 - 3x^2y^2 + 3xy^4 - y^6 +x^3 - 3x^2y^3 + 3xy^6 - y^9 = 0$$
Seeing patterns but I'm out of time for this morning, I'll post later if I get there.
