Analysing the roots of a ninth degree polynomial $$(x-3)^9+(x-3^2)^9+(x-3^3)^9.....+(x-3^9)^9=0$$ $\text{has} A)\text{all real roots} $ B)$3,3^2,..3,^9 C)\text{1 real  8 imaginary roots} D)\text{5 real 4 imaginary roots}$ .$$\text {my approach}$$ If we take it as $f(x)$now with some knowledge of calculus we can see differentiating it we get an expression which is always positive so the $f(x)$ is continuously increasing.But this question was asked in one of the class tests much before we were introduced to calculus. So is there any way where we can use general theory of equations or algebra to get the answer.Thanks!

If it were me in (a particularly inspired day of) early high school, I'd say there's no real need for derivatives: $f(x)$ is the sum of $9$ strictly increasing functions - a.k.a. $(x-3^m)^9$. So it is strictly increasing. Now, technically you need the Bolzano-Weierstrass argument, but I'd personally say that it is intuitive.
