xb⁻¹xb=xb⁻¹+b=x1/b+b=x(b²+1)/b.
If (b2+1)/b=0 then x(b²+1)/b=1. However, b2+1≠0 for any b. This also applies if x is replaced by a or c; or if b is replaced by a or c.
Note that neither a nor c appears in the proposed equality. The question has therefore been wrongly or misleadingly stated or is incomplete.
One possible simple proposition would be xaxbxc=1, because xa+b+c=x0=1.