12. The minimum value of $f(x)=x^{6}-6 x(x \in \mathbf{R})$ is $\qquad$
12. -5

Let $y=f(x)=x^{6}-6 x$, by $f(1)j(-x)$, so $y_{\operatorname{man}}$ is obtained when $x>0$.
At this time, $x^{5}+\frac{-y}{x^{2}}=6,-y \leqslant 0$ is possible, but to find $y_{\operatorname{mnn}}$, only consider $-y<0$, i.e., the situation where $y<0$.
At this time, $6=x^{5}+\frac{-y}{x}=x^{5}+\frac{-y}{5 x}+\frac{-y}{5 x}+\frac{-y}{5 x}+\frac{-y}{5 x}+\frac{-y}{5 x} \geqslant 6 \sqrt[6]{\left(\frac{-y}{5}\right)^{5}}$, so $\sqrt[6]{\left(\frac{-y}{5}\right)^{5}} \leqslant 1$. Therefore, $y \geqslant-5$, with equality if and only if $x^{5}=\frac{-y}{5 x}$, i.e., $x=1$. Hence, $y_{\text {nú }}=-5$.