While defining the Mandelstam variables for a process, we only take these three combinations of the momentum vectors: \begin{equation} (p_1 + p_2)^2, (p_1 - p_3)^2, (p_1 - p_4)^2 \end{equation} My question is, why we don't take other combinations of the momentum vectors, e.g. \begin{equation} (p_1 - p_2)^2, (p_1 + p_3)^2, (p_2 - p_3)^2 \end{equation} etc.
Because they are not independent. Expand the definitions
\begin{eqnarray}s&\equiv&(p_1+p_2)^2 = m_1^2+m_2^2+2p_1\cdot p_2\Longrightarrow 2p_1\cdot p_2=s-m_1^2-m_2^2,\\ t&\equiv&(p_1-p_3)^2=m_1^2+m_3^2-2p_1\cdot p_3\Longrightarrow 2p_1\cdot p_3=-t+m_1^2+m_3^2,\\ u&\equiv&(p_1-p_4)^2=m_1^2+m_4^2-2p_1\cdot p_4,\Longrightarrow 2p_1\cdot p_4=-u+m_1^2+m_4^2\end{eqnarray}
Then you see that $s,t,u$ are essentially equivalent to $p_1\cdot p_2$, $p_1\cdot p_3$ and $p_1\cdot p_4$. In that case your other combinations give
\begin{eqnarray}(p_1-p_2)^2&=&m_1^2+m_2^2-2p_1\cdot p_2=2m_1^2+2m_2^2-s,\\ (p_1+p_3)^2&=&m_1^2+m_3^2+2p_1\cdot p_3=2m_1^2+2m_2^2-t,\\ (p_1+p_4)^2&=&m_1^2+m_4^2+2p_1\cdot p_4=2m_1^2+2m_4^2-u.\end{eqnarray}
