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:
$$
\begin{align}
p(x^{12})-6 & = \big(x^{12}-1\big)+\big((x^{12})^2-1\big)+\big((x^{12})^3-1\big)+\big((x^{12})^4-1\big)+\big((x^{12})^5-1\big) \\
 & = (x^{12}-1)( \cdots ) = (x^6-1)(x^6+1)(\cdots) = p(x)\cdot(x-1)(x^6+1)(\cdots)
\end{align}
$$

[ EDIT ]  The above can also be derived directly from the more general:

$$a(x) - b(x) \mid p\big(a(x)\big) - p\big(b(x)\big) \implies x^{12} - 1 \mid p(x^{12})-p(1) \implies p(x) \mid p(x^{12}) - 6$$
