= Solution
<Linearity of Stokes flow> and <rotational symmetry>, including <reflection in mathematics> invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a <polar vector>; its linear dependence on a single <vector> $\mathbf A$ must be $c\mathbf A$. A <traceless second-rank tensor> $\mathbf B$ cannot produce a polar <vector> by an isotropic linear map: <tensor contraction> with the <identity matrix> vanishes and <tensor contraction> with the <totally antisymmetric tensor> vanishes because $\mathbf B$ is a <symmetric second-rank tensor>. <Angular velocity> is an <axial vector>. Neither $\mathbf A$ nor a symmetric $\mathbf B$ can produce one by an isotropic linear map. Products such as $\mathbf A\times\mathbf B\mathbf A$ are excluded by <Linearity of Stokes flow>. Thus $\boldsymbol\Omega=0$.
Use the <Unscaled Papkovich–Neuber representation>, with <harmonic functions> $\boldsymbol\Phi$ and $\chi$:
$$
\mathbf u=\nabla(\mathbf x\cdot\boldsymbol\Phi+\chi)-2\boldsymbol\Phi,
\qquad p-p_\infty=2\mu\nabla\cdot\boldsymbol\Phi.
$$
Indeed $\nabla\cdot\mathbf u=0$ and $\mu\nabla^2\mathbf u=\nabla p$ because each potential is a <harmonic function>. Put $r=|\mathbf x|$ and $S=\mathbf x\cdot\mathbf B\mathbf x$. The decaying <harmonic functions> with the necessary angular dependence are $\mathbf A\cdot\mathbf x/r^3$, $\mathbf B\mathbf x/r^3$ and $S/r^5$. The last is a <harmonic function> precisely because $\operatorname{tr}\mathbf B=0$. A $1/r$ <Stokeslet> is excluded by the <force-free> condition, and a <rotlet> by the <torque-free> condition. Try
$$
\boldsymbol\Phi=d\frac{\mathbf B\mathbf x}{r^3},\qquad
\chi=c\frac{\mathbf A\cdot\mathbf x}{r^3}+e\frac{S}{r^5}.
$$
The resulting <Stokes flow> is
$$
\mathbf u=c\left(\frac{\mathbf A}{r^3}-\frac{3(\mathbf A\cdot\mathbf x)\mathbf x}{r^5}\right)
-3d\frac{S\mathbf x}{r^5}
+e\left(\frac{2\mathbf B\mathbf x}{r^5}-\frac{5S\mathbf x}{r^7}\right).
$$
At $r=a$, the coefficient of $\mathbf n(\mathbf A\cdot\mathbf n)$ fixes $c=a^3/3$; the remaining constant <vector> gives $\mathbf U=-2\mathbf A/3$. In the $\mathbf B$ mode, matching $\mathbf B\mathbf n-(\mathbf n\cdot\mathbf B\mathbf n)\mathbf n$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete <two-mode tensorial squirmer flow> is
$$
\boxed{\begin{aligned}
\mathbf u(\mathbf x)&=\frac{a^3}{3r^3}\left[\mathbf A-3(\mathbf A\cdot\widehat{\mathbf x})\widehat{\mathbf x}\right]
\\&\quad+\frac{3a^2 S\mathbf x}{2r^5}
+\frac{a^4}{2}\left(\frac{2\mathbf B\mathbf x}{r^5}-\frac{5S\mathbf x}{r^7}\right),\\
p-p_\infty&=3\mu a^2\frac{S}{r^5},\qquad
\mathbf U=-\frac23\mathbf A,\quad\boldsymbol\Omega=0.
\end{aligned}}
$$
The <boundary condition> is satisfied in the laboratory frame, so the <Stokes flow> tends to zero at infinity. The <potential dipole> in the $\mathbf A$ mode gives an <irrotational flow> and decays as $r^{-3}$; the leading $\mathbf B$ mode is a <stresslet>, decaying as $r^{-2}$. Neither carries a net <force> or <torque>. Direct <surface integral> averaging with the <surface slip velocity> formula gives the same translation, providing an independent check. <Uniqueness of Stokes flow> then identifies this decaying solution.
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