\gamma.$$ Thus $$a = \lambda^3 a,~\alpha = \lambda^3 \alpha.$$ If both $a$ and $\alpha$ are zero, then $a=b=c=\alpha = \beta =  \gamma=0$, and $g=0$, a contradiction. Thus either $a \ne 0$ or $\alpha \ne 0$. In both cases $
      \lambda^3 = 1$. If $\lambda = 1$, then $a=b=c$ and $\alpha = \beta =  \gamma$, and hence $g$ is symmetric, contradicting Lemma \[AAA\]. Thus $\lambda \ne 1$ and $\lambda^3 = 1$. Therefore $\lambda$ is a primitive third root of 1 (and ${\mbox{char~}}k$ cannot be 3). Thus $g$ is of the form $$g = a (x^2+\lambda y^2+ \lambda^2 z^2) + \alpha (yz+\lambda zx+ \lambda^2 xy),$$ where $\lambda$ is a primitive third root of 1. Applying the permutation $y \mapsto z \mapsto y$ to $g$, we obtain another factor $$h = a (x^2+\lambda^2 y^2+ \lambda z^2) + \alpha (yz+\lambda^2 zx+ \lambda xy),$$ of $f$ that is not associate of $g$. Therefore $gh$ divides $f$, and has the same degree as $f$. Therefore $f = c gh$, where $c \in k$. Equating the coefficients of $x^3z$ and $x^2yz$ in the identity $f=cgh$, we obtain $-a\alpha c = 0$ and $2a\alpha  c - \alpha^2 c  = 0$. Thus $\alpha =0$ and $$g = a (x^2+\lambda 