Cyclotomic polynomial simplification I try to work out what $\Phi_{12}(z)$ is:
By the fundamental theorem of arithmetic:
$$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)(z-\xi^6)\\(z-\xi^7)(z-\xi^8)(z-\xi^9)(z-\xi^{10})(z-\xi^{11})$$
$$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)\\ 
(z+\xi^0)(z+\xi^1)(z+\xi^2)(z+\xi^3)(z+\xi^4)(z+\xi^5)
$$
$$\Phi_{12}(z)=(z^2-\xi^{0\cdot 2})(z^2-\xi^{1\cdot 2})(z^2-\xi^{2\cdot 2})(z^2-\xi^{3\cdot 2})(z^2-\xi^{4\cdot 2})(z^2-\xi^{5\cdot 2})$$
$$\Phi_{12}(z)=(z^2-\xi^{0})(z^2-\xi^{2})(z^2-\xi^{4})(z^2-\xi^{6})(z^2-\xi^{8})(z^2-\xi^{10})$$
$$\Phi_{12}(z)=(z^2-\xi^{0})(z^2-\xi^{2})(z^2-\xi^{4})(z^2+\xi^{0})(z^2+\xi^{2})(z^2+\xi^{4})$$
$$\Phi_{12}(z)=(z^4-\xi^{0\cdot 2})(z^4-\xi^{2\cdot 2})(z^4-\xi^{4\cdot 2})$$
$$\Phi_{12}(z)=(z^4-\xi^{0})(z^4-\xi^{4})(z^4-\xi^{8})$$
$$\Phi_{12}(z)=(z^4-\xi^{0})(z^8-z^4(\xi^{4}+\xi^{8})+\xi^{4}\xi^{8})$$
$$\Phi_{12}(z)=(z^4-1)(z^8-z^4+1)$$
Why is $\Phi _{12} (z)$ given as $z^4-z^2+1$?
(Or, more precisely, what is wrong with my method (or am I not noticing their equivalence)?).

I suppose $\xi$ is a primitive   $12^{th}$ root of unity.
The polynomial $\Phi_{12}(x)$ is the product $\prod(z-\xi^k)$ over the primitive $12^{th}$ roots of unity (i.e. for $k=1,5,7,11$). 
What you are calculating is, $\displaystyle{\prod_{k=0}^{11}(z-\xi^k)}$, the  product over all $12^{th}$ roots of unity which is just the polynomial $z^{12}-1$.
