= Solution
For side-by-side spheres, reflection in the vertical mid-plane exchanges the spheres while preserving their common velocity, so their axial drags are equal. For vertically aligned spheres, reflection in the horizontal mid-plane exchanges them and reverses velocity; applying linearity to reverse all velocities back restores the original problem and again gives equal axial drag. These are symmetries of the exact two-sphere Stokes problem.
At a distant side-by-side centre, the leading flow produced by one sphere is $3aU/(4b)$ in the direction of translation. On its axis it is $3aU/(2b)$. This ambient entrainment lowers the relative velocity of the other sphere. Using <Stokes drag law> with that relative velocity gives the <mutual drag reduction of two distant translating spheres>:
$$
\boxed{D_{\rm side}=6\pi\mu aU\left[1-\frac{3a}{4b}+O((a/b)^2)\right],}
$$
$$
\boxed{D_{\rm axial}=6\pi\mu aU\left[1-\frac{3a}{2b}+O((a/b)^2)\right].}
$$
The feedback of the corrected force into the other sphere's flow enters at the next order. If the coefficient $6\pi$ is needed directly from the preceding solution, its pressure anomaly is $p=3\mu Ua\cos\theta/(2R^2)$. At the surface the traction components are $t_R=-3\mu U\cos\theta/(2a)$ and $t_\theta=3\mu U\sin\theta/(2a)$; their axial projection is the constant $-3\mu U/(2a)$. Integrating over $4\pi a^2$ gives the single-sphere force $-6\pi\mu aU\mathbf e_z$, as used above.
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