10. Let $a, b, c \in \mathbf{R}$, and there exist $\alpha, \beta, \gamma \in\{-1,1\}$, such that $a \alpha + b \beta + c \gamma = 0$, find the minimum value of $f=\left(\frac{a^{3}+b^{3}+c^{3}}{a b c}\right)^{2}$.
10. Solution: $\because \quad \alpha, \beta, \gamma \in\{-1,1\}, \therefore$ by the pigeonhole principle, it is known that among the three real numbers $\alpha, \beta, \gamma$, at least two are equal. (1) All three real numbers are equal, $\alpha=\beta=\gamma \in\{-1,1\}$, then it must be that $a+b+c=0$, thus $f=\left(\frac{a^{3}+b^{3}-(a+b)^{3}}{a b c}\right)^{2}=\left(\frac{3 a b(a+b)}{a b c}\right)^{2}$ $=9$. (2) The three real numbers are not all equal, without loss of generality, let $\beta=\gamma \neq \alpha$, then it must be that $a=b+c$, thus $f=\left(\frac{(b+c)^{3}+b^{3}+c^{3}}{(b+c) b c}\right)^{2}=$ $\left(\frac{(b+c)^{2}+b^{2}-b c+c^{2}}{b c}\right)^{2}=\left(\frac{2\left(b^{2}+c^{2}\right)+b c}{b c}\right)^{2} . \therefore$ when $b c>0$, it further follows that $f \geqslant\left(\frac{4|b c|+b c}{b c}\right)^{2}=25, “=”$ holds when $b=c \neq 0$. In summary, $f_{\min }=9$.