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Suppose we have the following:
$E(\vec{k}) = E_0 - A(\cos(k_x a) + \cos(k_y a) + \cos(k_z a))$, a tight binding energy.

I know that the effective mass is given by $m^{*} = \hbar^2 [\frac{\partial^2 E}{\partial k_i \partial k_j}]^{-1}$ (where $i$ and $j$ denote k-space vector directions) which is just the curvature of the band.

In the current case as I have cosines of $k_x,k_y,k_z$

Does this means that I have three effective masses?
$$m^{*}_{xx} = -\hbar^2 \frac{1}{Aa^2 \cos(k_x a)}$$ $$m^{*}_{yy} = -\hbar^2 \frac{1}{Aa^2 \cos(k_y a)}$$ $$m^{*}_{zz} = -\hbar^2 \frac{1}{Aa^2 \cos(k_z a)}$$

Could we not come up with a single expression for the effective mass as the energy band is isotropic?

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  1. The band is indeed isotropic given OP's dispersion relation for the energy.
  2. The effective mass components are $$m^\star_{ij}=\hbar^2\left(\left.\frac{\partial^2E}{\partial k_i\partial k_j}\right|_{\color{red}{\vec{k}_0}}\right)^{-1}$$ I.e., the second derivatives must be evaluated at the point where the band has the minimum, in OP's case $\vec{k}_0=0$. With this, check that $m^\star_{xx}=m^\star_{yy}=m^\star_{zz}$.
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