= Solution
Under the stated linear field expansion, the <Bunch-Davies vacuum> is a centered Gaussian state. By <Wick theorem>, vacuum correlators reduce to pair contractions; an odd product has no complete pairing. In particular,
$$
\boxed{\langle0|\widehat{\delta\phi}(\tau,\mathbf x)\widehat{\delta\phi}(\tau,\mathbf y)\widehat{\delta\phi}(\tau,\mathbf z)|0\rangle=0.}
$$
This holds for arbitrary spatial arguments as a regulated correlator or distribution. One can also use number parity: $\mathcal P=(-1)^{\widehat N}$ leaves the vacuum unchanged and sends every creation or annihilation operator, hence the linear field $\widehat f$, to its negative. An odd product consequently has expectation equal to its own negative. Thus <vanishing odd correlators of a free Gaussian field> gives
$$
\boxed{\langle0|[\widehat f(\tau,\mathbf0)]^{2n+1}|0\rangle=0\qquad(n=0,1,2,\ldots).}
$$
For a coincident product, impose a momentum regulator or smear the field before taking the expectation; the zero result holds for every such symmetry-preserving regulator. It is not a claim that the unregulated local power is an ordinary finite random variable. Interactions, a non-Gaussian state, or inclusion of a nonzero background mean can invalidate the odd-moment conclusion. Within this free perturbation calculation, the <primordial bispectrum> vanishes.
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