Given $3$-vectors $A$ and $B$ and an unknown function $f:\mathbb{R^3}\to \mathbb{C}$, are there any solutions to this functional equation? $$f(A)f(B)=f(A+B+A\times B)$$ apart from the trivial solutions $f(A)=1$ or $f(A)=0$.
Edit: As a simpler example a solution to $f(A)f(B)=f(A+B)$ is $f(A)=e^{k\cdot A}$ for some $3$-vector $k$.
The functional equation has no non-trivial solution.
For any $a, b \in \mathbb{R}^3$ such that $|a| = |b|$ and $a \ne -b$. Let
$$A = \frac{a+b}{4} + \frac{2}{|a+b|^2}(a \times b) \quad\text{ and }\quad B = \frac{a+b}{4} - \frac{2}{|a+b|^2}(a \times b) $$ Notice $$\begin{align}A + B + A \times B &= \frac{a+b}{2} - \frac12(A+B)\times(A-B)\\ &= \frac{a+b}{2} - \frac{1}{|a+b|^2} (a+b)\times(a \times b)\\ &= \frac{a+b}{2} - \frac{1}{|a+b|^2}\left((a\cdot b + |b|^2)a - (|a|^2+b\cdot a)b\right)\\ &= \frac{a+b}{2} - \frac{a-b}{2}\\ &= b \end{align} $$ and similary, $A +B - A \times B = a$. We have
$$\begin{align} f(a) &= f(A+B-A\times B) = f(A+B+B\times A) = f(B)f(A)\\ f(b) &= f(A+B+A\times B) = f(A)f(B)\\ \implies f(a) &= f(b) \end{align} $$ For any $a \in \mathbb{R}^3$ such that $a \ne 0$, take a $b\in \mathbb{R}^3$ with $|b| = |a|$ and $b \ne a, -a$. This leads to
$$f(a) = f(b) = f(-a)$$ Combine these, we can deduce $f(a)$ depends only on $|a|$.
For any $a \in \mathbb{R}^3$, we have
$$f(a)^2 = f(a)f(-a) = f(a - a - a \times a) = f(0)$$
This forces $f(a)$ to be a constant function.
Since $f(0)^2 = f(0+0+0\times 0) = f(0)$, the constant can only be $0$ or $1$.
i.e. the functional equation has no non-trivial solution.
