Suppose we have a rank-3 tensor $T$ on some vector space $\mathbb{V}$. We can view $T$ as a map: $$T: \mathbb{V} \times \mathbb{V} \times \mathbb{V} \to \mathbb{R},$$ which maps triples of vectors in $\mathbb{V}$ to the real numbers. By the definition of tensors, $T$ is linear: $$ T(\alpha \vec{v}_1 + \beta \vec{v}_2, \vec{u}, \vec{w}) = \alpha T(\vec{v}_1 , \vec{u}, \vec{w}) + \beta T(\vec{v}_2 , \vec{u}, \vec{w}), $$ and similarly for the other two arguments.
Suppose, further, that $T$ is completely symmetric in its inputs: $$ T(\vec{v}_1, \vec{v}_2, \vec{v}_3) = T(\vec{v}_1, \vec{v}_3, \vec{v}_2) = T(\vec{v}_2, \vec{v}_1, \vec{v}_3) = \dots $$ for any $\{\vec{v}_1, \vec{v}_2, \vec{v}_3\} \in \mathbb{V}$, and for all six permutations of the inputs.
Finally, suppose that we know that $T(\vec{v}, \vec{v}, \vec{v}) = 0$ for all $\vec{v} \in \mathbb{V}$.
My question:
Does it follow that $T(\vec{v}_1, \vec{v}_2, \vec{v}_3) = 0$ for all $\{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$?
The analogous statement is true for rank-2 tensors, so it would seem strange to me if it wasn't true here as well. On the other hand, the argument I'm familiar with for the rank-2 case [i.e.: plug $\vec{v}_1 + \vec{v}_2$ into both slots and use the linearity properties to argue that $T(\vec{v}_1, \vec{v}_2) = 0$] doesn't carry over to this case, so I'm not sure how to prove it.
Expanding by linearity and removing all terms we know are zero, and using symmetry to gather the rest of the terms, we get $$\begin{align} 0 = &\,\frac13T(\vec v_1 + \vec v_2 + \vec v_3,\vec v_1 + \vec v_2 + \vec v_3, \vec v_1 + \vec v_2 + \vec v_3)\\ =&\, T(\vec v_1, \vec v_1, \vec v_2)+ T(\vec v_1, \vec v_2, \vec v_2) \\ &+ T(\vec v_1, \vec v_1, \vec v_3) + T(\vec v_1, \vec v_3, \vec v_3) \\ &+ T(\vec v_2, \vec v_2, \vec v_3) + T(\vec v_2, \vec v_3, \vec v_3)\\& + 2T(\vec v_1, \vec v_2, \vec v_3)\end{align} $$ (where the $\frac13$ is just for readability). In order to get any further, we need to study the value of $T(\vec v, \vec v, \vec u) + T(\vec v ,\vec u , \vec u)$, which appears in three different variations above. Applying $T$ to three copies of $\vec v + \vec u$, we get $$ 0 = \frac13T(\vec v + \vec u, \vec v + \vec u, \vec v + \vec u)= T(\vec v, \vec v, \vec u) + T(\vec v, \vec u, \vec u) $$ and therefore we conclude that $T$ vanishes.