  We give sufficient conditions on a symmetric tensor S in S^dF^n to satisfy
the equality: the symmetric rank of S, denoted as srank(S), is equal to the
rank of S, denoted as rank(S). This is done by considering the rank of the
unfolded S viewed as a matrix A(S). The condition is: rank(S) is in
{rank(A(S)),rank (A(S))+1}. In particular, srank(S)=rank(S) for S in S^dC^n for
the cases (d,n) in {(3,2),(4,2),(3,3)}. We discuss the analogs of the above
results for border rank and best approximations of symmetric tensors.
