Let $P(x) =1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P(x^{12})$ is divided by $P(x)$? [closed]

Let $P(x) =1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P(x^{12})$ is divided by $P(x)$?

closed as off-topic by heropup, user223391, Pragabhava, Watson, Stefan4024Sep 22 '16 at 22:29

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• should there be exponents? – Jorge Fernández Hidalgo Sep 22 '16 at 16:36
• When you work with higher degree polynomial and carry out multiplication and division you are not doing Linear Algebra. – P Vanchinathan Sep 22 '16 at 16:39
• – Jyrki Lahtonen Sep 22 '16 at 17:09
• Please improve your question by mentioning your attempts. There is an interesting interpolation approach related with the sixth roots of unity. – Jack D'Aurizio Sep 23 '16 at 0:45
• In particular, the remainder is $\color{red}{6}$ because the value of $Q(x)=P(x^{12})$ at any sixth root of unity is six. – Jack D'Aurizio Sep 23 '16 at 0:53

HINT: $$P(x) = \frac{x^6 - 1}{x-1}, \quad \text{when} \quad x \not = 1$$
• Secondary hint: $a^2-b^2$ – Hagen von Eitzen Sep 24 '16 at 9:24