. For what value of $k$ do we have the identity (true for all $a, b, c$):

$$
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k a b c ?
$$
. It suffices to develop each of the two members. All terms are identical except the terms in $a b c$ which appear twice on the left and $3+k$ times on the right. Therefore, $k=-1$. We note that for $a=b=c=1$, we get: $8=9+k$, so the only possible value of $k$ is -1, but this does not dispense with doing the calculation to prove the identity when $k=-1$.