# Find irreducible factor of the polynomial

Find all irreducible factors of the polynomial $$x^{15}-1$$ over $$\Bbb{Q}$$.

Clearly, this is a cyclotomic polynomial.

I can find all the irreducible factors but that is very calculative:

\begin{align} \Phi_1(x)&=x-1\\ \Phi_3(x)&=\frac{x^3-1}{\Phi_1(x)}\\ \Phi_5(x)&=\frac{x^5-1}{\Phi_1(x)}\\ \Phi_{15}(x)&=\frac{x^{15}-1}{\Phi_1(x)\Phi_3(x)\Phi_5(x)}. \end{align} The last equation is very calculative. That's why I am looking for any less time consuming method.

Thanks!

• HINT: $\left ( x^5 \right )^{3} - 1^{3}$, what lovely formula do you know for this difference? Nov 17 '20 at 19:50
• @Fede1 I tried it alreday. but it is not useful Nov 17 '20 at 19:50
• $\Phi_5(x)=\frac{x^5-1}{\Phi_1(x)\Phi_3(x)},$ is wrong Nov 17 '20 at 19:56
• @mathcounterexamples.net Yes, Edited Nov 17 '20 at 19:58

You have $$\Phi_1(x) \Phi_5(x)= x^5-1$$
$$\Phi_{15}(x)= \frac{x^{15}-1}{(x^5-1) \Phi_3(x)}= \frac{(x^5)^3-1}{(x^5-1)\Phi_3(x)}=\frac{x^{10}+x^5+1}{\Phi_3(x) }$$
• $\Phi_{15}(x)=x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$. Thanks!(+1) Nov 17 '20 at 20:22