A bilinear form $B(x,y)$ satifies $B(a,b)B(c,a)=B(b,a)B(a,c)$ implies it is either symmetric or skew-symmetric. Question:
If a bilinear form $B(x,y)$ satisfies for $\forall a,b,c \in V_{\Bbb{F}}$$$B(a,b)B(c,a)=B(b,a)B(a,c)$$Then $B(x,y)$ is either symmetric or skew-symmetric.

If I let $c=a+b$, then get $B(a,a)\left[B(a,b)-B(b,a)\right]=0\cdots(*)$. If $\text{char}(\Bbb{F})\not=2$, then $B(a,a)=0\cdots(1)$ for $\forall a \in V_{\Bbb{F}}$ $\Longleftrightarrow$$B(x,y)$ is skew-symmetric. and $B(a,b)-B(b,a)=0\cdots(2)$ for $\forall a,b\in V_{\Bbb{F}}$ means $B(x,y)$ is symmetric. But I can't deduce$(1)$ and $(2)$ from $(*)$. Could you help me?
 A: I don't know how to fix your argument, but from a matrix theoretic persective, the proof is rather straightforward (despite a bit long). For convenience, I write $V$ instead of $V_{\mathbb{F}}$.
It suffices to prove the statement for the finite dimensional case, for,
if $B$ is neither symmetric nor skew-symmetric, then there exist
$u,v,x,y\in V$ such that $B(u,v)\ne B(v,u)$ and $B(x,y)\ne-B(y,x)$;
hence $B$ is neither symmetric nor skew-symmetric on the finite (actually, at most four) dimensional subspace spanned by $u,v,x,y$.
Suppose $V$ is finite dimensional. Then we can identify $B$ with its
matrix representation, and
\begin{align*}
B(a,b)B(c,a) \equiv B(b,a)B(a,c)
&\Leftrightarrow B(c,a)B(a,b) \equiv B(b,a)B(a,c)\\
&\Leftrightarrow c^\top Baa^\top Bb \equiv b^\top Baa^\top Bc\\
&\Leftrightarrow Baa^\top B \equiv (Baa^\top B)^\top\\
&\Leftrightarrow BSB=B^\top SB^\top \text{ for every symmetric matrix } S.\tag{1}
\end{align*}
If $\operatorname{char}(F)=2$, by considering every $S$ with a symmetric pair of off-diagonal ones and zeros elsewhere, $(1)$ immediately implies that $B$ is symmetric (hence also skew-symmetric).
Suppose $\operatorname{char}(F)\ne 2$. We can rewrite $(1)$ as
$$KSH+HSK = 0 \text{ for all symmetric matrix } S,\tag{2}$$
where $H$ and $K$ denote respectively the symmetric part and skew-symmetric part of $B$ (the decomposition of $B$ into $H+K$ is possible because $\operatorname{char}(F)\ne 2$). By applying a congruence transform $H\leftarrow P^\top HP,\ K\leftarrow P^\top KP$, we may assume that $H$ is a block-diagonal matrix of the form
$$
\pmatrix{\widehat{H}&0\\ 0&0_{m\times m}},
$$
where $m$ is the nullity of the linear map $x\mapsto Hx$ and $\widehat{H}$ is symmetric and invertible.
It follows from $(2)$ that in the same partitioning, all subblocks of $K$, except the leading principal one, are zero. So,


*

*if $H\ne0$, we can put $S=\operatorname{diag}(\widehat{H}^{-1},\ 0_{m\times m})$ and obtain $2K=0$, i.e. $B$ is symmetric;

*if $H=0$, then $B$ is skew-symmetric.

