Algebra with three anti-commutator relations

Question

Let $u,v,w \in \mathbb{F}_p^{\times}$. Consider the $\mathbb{F}_p$-algebra $V$ generated by $ a,b,c$ and the relations $$u a^2 = bc + cb$$ $$v b^2 = ac + ca$$ $$w c^2 = ab + ba$$ Is $V$ generated by the $a^i b^j c^k$ with $i,j,k \geq 0$ as an $\mathbb{F}_p$-module?

It is enough to show that $b a^i b^j$ and $c a^i b^j$ lie in the span. I tried induction on $i+j$, but I am going in circles. Even for $i+j=2$ already. Maybe it is not true actually? The corresponding question for $2$ generators has a positive answer, though.

Also, does $V$ have a name, or does $V$ belong to a more well-known class of algebras?

Background: I want to understand the nilpotent elements of $A \otimes_{\mathbb{F}_p} \mathbb{F}_{p^3}$, where $A$ is a certain non-commutative $\mathbb{F}_p$-algebra. When there is an irreducible binomial $T^3 - \lambda \in \mathbb{F}_p[T]$ (this is the case when $p \equiv 1 \bmod 3$), we get equations of the type above for any element $a+bT+cT^2$ of $A \otimes_{\mathbb{F}_p} \mathbb{F}_{p^3} = A[T]/\langle T^3 - \lambda \rangle$ that squares to $0$.

Answer 1

I ran a quick test on magma, asking it the dimension of the algebra defined by your relations with (chosen pretty much at random) $p=5$, $u=2$, $v=3$, $w=4$, and with the extra relations $a^3=0$, $b^3=0$, $c^n=0$, and let $n$ run from $3$ to $30$. The answer was consistent in each case with the statement that the dimension of the resulting algebra is $20n-24$, and therefore inconsistent with your statement, which predicts dimension at most $9n$. So the answer is "no".