A complex sum!

Rewrite the sum as:

$\displaystyle \sum_{k=0}^{22} \frac{1}{\alpha^{2k}+\alpha^k+1}=\sum_{k=0}^{22} \frac{1}{(\alpha^k-\omega)(\alpha^k-\omega^2)}=\frac{1}{\omega-\omega^2}\left(\sum_{k=0}^{22} \frac{1}{\alpha^k-\omega}-\sum_{k=0}^{22} \frac{1}{\alpha^k-\omega^2}\right)$

where $\omega$ and $\omega^2$ are non-real cube roots of unity.

Also,

$\displaystyle x^{23}-1=\prod_{i=0}^{22} \left(x-\alpha^k\right) \Rightarrow \ln\left(x^{23}-1\right)=\sum_{k=0}^{22} \ln\left(x-\alpha^{k}\right)$

Differentiate both the sides with respect to $x$ to obtain,

$\displaystyle \sum_{k=0}^{22} \frac{1}{x-\alpha^k}=\frac{23x^{22}}{x^{23}-1} \Rightarrow \sum_{k=0}^{22} \frac{1}{\alpha^k-x}=\frac{23x^{22}}{1-x^{23}}$

Substitute $x=\omega$ and $x=\omega^2$ to obtain the sums, i.e

$\displaystyle \sum_{k=0}^{22} \frac{1}{\alpha^k-\omega}=\frac{23\omega^{22}}{1-\omega^{23}}=\frac{23\omega}{1-\omega^2}$

$\displaystyle \sum_{k=0}^{22} \frac{1}{\alpha^k-\omega^2}=\frac{23\omega^{44}}{1-x^{46}}=\frac{23\omega^2}{1-\omega}$

Hence,

$\displaystyle \sum_{k=0}^{22} \frac{1}{\alpha^{2k}+\alpha^k+1}=\frac{1}{\omega-\omega^2}\left(\frac{23\omega}{1-\omega^2}-\frac{23\omega^2}{1-\omega} \right) =\boxed{\dfrac{46}{3}}$

NOTE : The sums can be evaluated without Calculus. See Calvin's comment on Jatin's solution here: Summing roots of unity .