If $p(x)=1+x+x^2+x^3+x^4+x^5$ find the remainder when $p(x^{12})$ is divided by $p(x)$ 
Question:
If $p(x)=1+x+x^2+x^3+x^4+x^5$ find the remainder when $p(x^{12})$ is divided by $p(x)$
Options:
(A) $0$
(B) $6$
(C) $1+x$
(D) $1+x+x^2+x^3+x^4$

My attempt:
Since $p(x)=1+x+x^2+x^3+x^4+x^5$ is a geometric series assuming $x<1$ with sum
$$S_1=\frac{1-x^6}{1-x}$$
and $p\left(x^{12}\right)$ also is a geometric series with sum
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x}$$
then
$$\frac{S_2}{S_1}= \frac{1-\left(x^{12}\right)^6}{1-x^6}$$
Now I can't seem to find a way to proceed further.
EDIT
I just found a mistake I made.
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x^{12}}$$
And not
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x}$$
So my division was wrong and so was my solution thereafter.
Sorry !

Hint: $\dfrac{S2}{S1}= \dfrac{1-(x^{12})^6}{1-x^6}= \dfrac{1-(x^6)^{12}}{1-x^6}= \dfrac{1-y^{12}}{1-y}$, for $y=x^6$.
