<s>[INST] How to find the remainder $f(x)$=$1+x+x^2+x^3+x^4+x^5$
What is the remainder when $f(x^{12})$ is divided by $f(x)$?
I think remainder theorem cannot be applied (at least directly) here.
I find that $f(-1^{12})$ - $f(-1)$ is $6$ but $f(1^{12})$ - $f(1)$ is $0$
 [/INST] Hint: It is easier to calculate the remainder of $f(x^{12})(x-1)$ when divided by $f(x)(x-1)=x^6-1$, then divide that remainder by $x-1$ at the end.
Or more directly, note that:
$$x^6\equiv 1\pmod {f(x)}$$
so:
$$x^{12}\equiv 1\pmod{f(x)}$$
So what is $f(x^{12})\pmod{f(x)}$?
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