bXc = b -> b=0 with regularity but not infinity

["MoeBlee"]

Here's one that has me stumped:

In ZR(-I) (that's Z set theory with the axiom of regularity but not the
axiom of infinity), prove:

bXc = b -> b=0

where 'X' stands for the Cartesian product.

You may not use the 'infinite sequence of nested members' version of
regularity. But the following previous theorems are available (where
'e' stands for 'is a member of') along with the usual "first chapter"
theorems of set theory:

~SeS

~SeTeS

~SeTeVeS

S subset SXS -> S=0

Source: 'Axiomatic Set Theory' by Suppes, pg. 56, exercise 2 (Dover
edition).