If some tensor $T=(t\_{ijk})$ has that all of its (2 dimensional) slices (along all 3 axes) are of rank-1, does it follow that the tensor is also rank-1? That is, can be written as
$$ t\_{ijk}=a\_i b\_j c\_k$$
? 



Rank of order-3 tensor with all slices being rank-1
Let $t$ be a nonzero tensor. Then some $t\_{ijk}$ are nonzero, without loss of generality let $t\_{111}\neq 0$. Rescaling our tensor, we may assume that $t\_{111}=1$. Put $a\_i = t\_{i11}$, $b\_j = t\_{1j1}$ and $c\_k = t\_{11k}$. Then $t$ is rank $1$ if and only if $t\_{ijk} = a\_i b\_j c\_k$. 


If all $t\_{11k}$ are nonzero, then this follows from the condition on $2$-dimensional slices:
$$(t\_{ijk} t\_{11k}) t\_{111}^2 = (t\_{i1k} t\_{1jk}) t\_{111}^2 = (t\_{i1k} t\_{111}) (t\_{1jk} t\_{111}) = (t\_{i11} t\_{11k}) (t\_{1j1} t\_{11k}) = (t\_{i11} t\_{1j1} t\_{11k}) t\_{11k} $$
so
$$t\_{ijk} t\_{11k} = a\_i b\_j c\_k t\_{11k}.$$
One could obviously use conditions that various $t\_{1j1}$ or $t\_{i11}$ are nonzero instead.


Without any nonvanishing assumptions, the easiest thing I can see to do is to impose a condition on slanted $2 \times 2$ blocks: $t\_{i\_1 j\_1 k\_1} t\_{i\_2 j\_2 k\_2} = t\_{i\_1 j\_1 k\_2} t\_{i\_2 j\_2 k\_1}$ and likewise in switching the $i$-indices and the $j$-indices.


