  This paper shows that one cannot "hear" the rational cohomology ring of a
hyperbolic 3-manifold. More precisely, while it is well-known that strongly
isospectral manifolds have the same cohomology as vector spaces, we give an
example of compact hyperbolic 3-manifolds that are strongly isospectral but
have nonisomorphic rational cohomology rings. Along the way we implement a
computer program which finds the nullity of the cup product map
$H^1(M;\mathbb{Q})\wedge H^1(M;\mathbb{Q})\rightarrow H^2(M;\mathbb{Q})$ for
any aspherical space in terms of the presentation of the fundamental group.
