= Solution
{c}
\b[True.] Two applications of <Green's identities>, or two <integrations by parts>, give
$$
\iiint_Vu\Delta^2u\,dV
=\iiint_V(\Delta u)^2\,dV
+\iint_{\partial V}\bigl(u\partial_n\Delta u-\Delta u\partial_nu\bigr)\,dS.
$$
The <biharmonic equation> makes the left side zero, while the two <clamped boundary conditions> $u=\partial_nu=0$ annihilate the <boundary integral>. Consequently
$$
\iiint_V(\Delta u)^2\,dV=0,
$$
so $\Delta u=0$. Apply the first of <Green's identities> once more with $\phi=\psi=u$: because $u=0$ on the <boundary>, it gives $\int_V|\nabla u|^2=0$. By <path connectedness>, $u$ is constant, and the zero <boundary value> implies $\boxed{u=0}$. This argument assumes the classical differentiability needed for the displayed <integrations by parts>.
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