= Solution
Work in the frame in which the distant fluid is stationary, and take the <pressure> at infinity as zero. In the <Unscaled Papkovich–Neuber representation>, choose the harmonic potentials
$$
\boldsymbol\Phi=-\frac{3a}{4r}\mathbf V,\qquad
\chi=\frac{a^3}{4}\frac{\mathbf V\cdot\mathbf x}{r^3}.
$$
Both are harmonic away from the sphere's centre: $1/r$ is harmonic there and $\mathbf V\cdot\mathbf x/r^3=-\mathbf V\cdot\nabla(1/r)$. Substitution into $\mathbf u=\nabla(\mathbf x\cdot\boldsymbol\Phi+\chi)-2\boldsymbol\Phi$, $p=2\mu\nabla\cdot\boldsymbol\Phi$, gives the <translating sphere in Stokes flow>:
$$
\boxed{\mathbf u=\left[\frac{3a}{4r}(\mathbf I+\mathbf n\mathbf n)+\frac{a^3}{4r^3}(\mathbf I-3\mathbf n\mathbf n)\right]\mathbf V,\qquad
p=\frac{3\mu a}{2r^2}\mathbf V\cdot\mathbf n.}
$$
Here $\mathbf n=\mathbf x/r$. At $r=a$ the coefficients of $\mathbf I$ sum to one and those of $\mathbf n\mathbf n$ sum to zero, so the <no-slip boundary condition> is satisfied. The field decays at infinity. The <Papkovich–Neuber representation> enforces the <Stokes equation> and <incompressibility>, and <Uniqueness of Stokes flow> identifies the physical solution.
Differentiate radially at fixed direction $\mathbf n$:
$$
\left.\partial_r\mathbf u\right|_{a}
=\left[-\frac{3}{4a}(\mathbf I+\mathbf n\mathbf n)-\frac{3}{4a}(\mathbf I-3\mathbf n\mathbf n)\right]\mathbf V
=\boxed{\frac{3}{2a}(\mathbf n\mathbf n-\mathbf I)\mathbf V.}
$$
On the surface, multiplying the specified <Newtonian fluid stress tensor> by the outward normal from the solid gives
$$
\boldsymbol\sigma\mathbf n=\frac{3\mu}{2a}\left[(\mathbf V\cdot\mathbf n)\mathbf n-\mathbf V-\mathbf n(\mathbf V\cdot\mathbf n)\right]
=-\frac{3\mu}{2a}\mathbf V.
$$
Thus the <surface traction in translating-sphere Stokes flow> is uniform and the fluid's <force> on the sphere is
$$
\boxed{\mathbf F_{\rm fluid}=\int_{r=a}\boldsymbol\sigma\mathbf n\,dS=-6\pi\mu a\mathbf V.}
$$
This is the <Stokes drag law>; an external <force> of the opposite sign maintains the motion.
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