How can I demonstrate that $x-x^9$ is divisible by 30? How can I demonstrate that $x-x^9$ is divisible by $30$ whenever $x$ is an integer?
I know that $$x-x^9=x(1-x^8)=x(1-x^4)(1+x^4)=x(1-x^2)(1+x^2)(1+x^4)$$
but I don't know how to demonstrate that this number is divisible by $30$.

Let's factor $x^9-x$ like you have done:
$$
x^9-x=(x-1)x(x+1)(x^2+1)(x^4+1).\tag{$*$}
$$
Let's look at the RHS. The product of the first 2 terms is divisible by $2$ because it consists of 2 consecutive integers. Similarly, the product of the first 3 terms is divisible by $3$. Now, if you had
$$
(x-2)(x-1)x(x+1)(x+2)
$$
then of course that would be divisible by $5$ as well. But note this
$$
(x-1)x(x+1)(x^2+1)-(x-2)(x-1)x(x+1)(x+2)=5x(x^2-1)\equiv 0\pmod{5}.
$$
So the product of the first 4 terms of the RHS of ($*$) is also divisible by $5$. Now you're done.
