Comment upon nature of the roots

Question

How many roots are there of the following polynomial? How many are real, and how many are complex?

$$\left(x-3\right)^9+\left(x-3^2\right)^9+\left(x-3^3\right)^9+\left(x-3^4\right)^9+...+\left(x-3^9\right)^9=0$$

Answer 1

Let $k$ be a natural number and consider $a_1,a_2,\dots,a_n\in\mathbb{R}$ pairwise distinct. The function $$ f(x)=(x-a_1)^{2k+1}+(x-a_2)^{2k+1}+\dots+(x-a_n)^{2k+1} $$ has only one real root, that is, $c\in\mathbb{R}$ such that $f(c)=0$.

Indeed, its derivative is $$ f'(x)=(2k+1)\bigl((x-a_1)^{2k}+(x-a_2)^{2k}+\dots+(x-a_n)^{2k}\bigr) $$ which is positive. The intermediate value theorem will end the job.

Answer 2

You could write it this way, isn't it?
$\sum_{n=1}^{n=9}(x-3^n)^9$

I would try to simplify your equation, starting with the values you already know (the 3^n).
You would get $(x-3)^9 + (x-9)^9 + (x-27)^9 +...+(x-19683)^9 = 0$

Doesn't bring you the answer, but maybe a step closer.

Answer 3

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.