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 February 23rd, 2016, 12:22 PM #1 Member   Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 Hard factorization problem How can you factor $\displaystyle k^{10}+k^5+1$ to this beautiful form: $\displaystyle (k^2+k+1)(k^8-k^7+k^5-k^4+k^3-k+1)$ Last edited by greg1313; February 23rd, 2016 at 12:48 PM. Reason: Corrected latex for k^10 February 23rd, 2016, 04:36 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 \displaystyle \begin{align*}k^{10} + k^5 + 1 &= \frac{k^{15} - 1}{k^3 - 1}\cdot \frac{k^3 - 1}{k^5 - 1} \\ &= (k^{12} + k^9 + k^6 + k^3 + 1)\cdot \frac{k^2 + k + 1}{k^4 + k^3 + k^2 + k + 1} \\ &= (k^8 - k^7 + k^5 - k^4 + k^3 - k + 1)(k^2 + k + 1)\end{align*} The longer factor = $\displaystyle (k^2 - 2\cos(24°)k + 1)(k^2 - 2\cos(48°)k + 1)(k^2 - 2\cos(96°)k + 1)(k^2 - 2\cos(192°)k + 1)$. Thanks from topsquark Tags factorization, hard, problem Search tags for this page

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