= Solution
Let $R(t)$ rotate <vectors> from the rotating frame to the inertial frame. For any rotating-frame <vector> $\mathbf a$, differentiating $R\mathbf a$ adds $R(\boldsymbol\Omega\times\mathbf a)$ to $R\dot{\mathbf a}$. Thus an inertial particle <velocity> is $R(\mathbf u+\boldsymbol\Omega\times\mathbf x)$, and, for constant <angular velocity>, its <acceleration> is
$$
R\left[\frac{D\mathbf u}{Dt}+2\boldsymbol\Omega\times\mathbf u+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf x)\right].
$$
Insert this into the inertial <Euler equations> and express <forces> in the rotating coordinates. This gives
$$
\boxed{\frac{D\mathbf u}{Dt}+2\boldsymbol\Omega\times\mathbf u+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf x)=-\rho^{-1}\nabla p+\mathbf g.}
$$
The additional terms are the <Coriolis acceleration> and the centripetal term associated with frame rotation. There is no angular-<acceleration> term because $\boldsymbol\Omega$ is constant.
Write the conservative <body force> as $\mathbf g=-\nabla\Phi_g$. The supplied <vector> identity gives $\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf x)=-\tfrac12\nabla|\boldsymbol\Omega\times\mathbf x|^2$. For constant <density>, define
$$
\boxed{P=p+\rho\Phi_g-\frac\rho2|\boldsymbol\Omega\times\mathbf x|^2.}
$$
Moving the centrifugal and conservative terms to the right then gives $\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u+2\boldsymbol\Omega\times\mathbf u=-\rho^{-1}\nabla P$. Only the <gradient> of this reduced <pressure> matters, so an additive spatial constant is immaterial.
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