Put
$$
S_1=\sum_{i=1}^3p_i,\qquad S_2=\sum_{i=1}^3p_i^2.
$$
Contracting the curvature components gives
$$
R_{00}=\frac{S_1-S_2}{t^2},
\qquad
R_{ii}=\frac{p_i(S_1-1)}{t^2}
$$
in the orthonormal frame, with no sum on $i$. Vacuum requires $S_1=S_2$ and $p_i(S_1-1)=0$ for every $i$. Since the $p_i$ are not all zero, some $p_i\neq0$, forcing $S_1=1$ and then $S_2=1$. Conversely these conditions make every Ricci component vanish. Thus the <Kasner metric> is vacuum exactly when
$$
\boxed{\sum_i p_i=1,\qquad \sum_i p_i^2=1},
$$
so $A=B=1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 / 3 / e / Solution

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The Kasner metric is the anisotropic vacuum cosmology
$$
ds^2=-dt^2+\sum_{i=1}^3t^{2p_i}(dx^i)^2,
$$
where the vacuum Einstein equations require $\sum_i p_i=\sum_i p_i^2=1$.


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