You can observe that you can pair $BC$ in the first item with $BC$ in the fourth one:
However, you could also pair $AC$ in the second item with $AC$ in the fourth one:
Similarly, pairing $AB$ in the third item with $AB$ in the fourth one:
Note that none of these final expressions is symmetric in $A$, $B$ and $C$, whereas the original one is.
This gives the idea of doing the three pairings all together, which is obtained by adding $ABC$ twice, justified because $P+P=P$:
Alternatively, we get a symmetric expression by summing up the three we got: