= Solution
The undisturbed ambient solution is smooth and free of <body forces> throughout the ball $r\le a$. Its <Stokes equations> imply
$$
\mu\nabla^2\mathbf u_\infty=\nabla p_\infty,\qquad
\nabla\cdot\mathbf u_\infty=0.
$$
Taking the divergence gives $\nabla^2p_\infty=0$, and a further <Laplacian> therefore gives
$$
\boxed{\nabla^4\mathbf u_\infty=\frac1\mu\nabla(\nabla^2p_\infty)=0.}
$$
Each <velocity> component obeys the <biharmonic equation> in the whole ball.
Expand a component of $\mathbf u_\infty$ at the centre and average over $\mathbf x=a\mathbf n$. The odd spherical moments vanish, while
$$
\langle n_i n_j\rangle=\frac13\delta_{ij},\qquad
\langle n_i n_j n_k n_l\rangle
=\frac1{15}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}).
$$
Thus the <spherical mean of a biharmonic function> gives
$$
\langle\mathbf u_\infty\rangle_{S_a}
=\left[1+\frac{a^2}{6}\nabla^2+\frac{a^4}{120}\nabla^4+\cdots\right]
\mathbf u_\infty(0)
=\mathbf u_\infty(0)+\frac{a^2}{6}\nabla^2\mathbf u_\infty(0).
$$
The truncation is exact, not just a small-sphere approximation: all terms containing two or more Laplacians vanish. One can avoid any Taylor-convergence issue by letting $M(r)=\langle\mathbf u_\infty(r\mathbf n)\rangle$. Since $\nabla^2\mathbf u_\infty$ is harmonic, its <spherical mean> is $\nabla^2\mathbf u_\infty(0)$, so
$$
M''+\frac2rM'=\nabla^2\mathbf u_\infty(0),\qquad M(0)=\mathbf u_\infty(0),\quad M'(0)=0.
$$
Integrating gives $M(r)=\mathbf u_\infty(0)+r^2\nabla^2\mathbf u_\infty(0)/6$ for every admissible $r$.
Combining with the force-free result of part <2/c> proves <Faxén's first law> in the requested form:
$$
\boxed{\mathbf V=\left.\left(1+\frac{a^2}{6}\nabla^2\right)\mathbf u_\infty\right|_{\mathbf x=0}.}
$$
The force-free qualification carries over. Local absence of ambient <body forces> and regularity inside the ball are also needed; if forcing or a singularity lies inside it, the harmonic-pressure and biharmonic-mean argument does not apply unchanged.
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