6. If $0^{\circ}<\alpha<30^{\circ}$, and $\sin ^{6} \alpha+\cos ^{6} \alpha=\frac{7}{12}$, then $1998 \cos \alpha=$

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6. $333 \sqrt{30}$.
$\sin ^{6} \alpha+\cos ^{6} \alpha=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{4} \alpha-\sin ^{2} \alpha \cos ^{2} \alpha+\cos ^{4} \alpha\right)=1-3 \sin ^{2} \alpha \cos ^{2} \alpha=\frac{7}{12}$, so $\sin ^{2} \alpha \cos ^{2} \alpha=\frac{5}{36}$, then $\left(\sin ^{2} \alpha\right)^{2}=\frac{20}{36}=\frac{5}{9}$. Since $\alpha \in\left(0, \frac{\pi}{6}\right)$, hence $2 \alpha \in\left(0, \frac{\pi}{3}\right)$, then $\sin 2 \alpha=\frac{\sqrt{5}}{3}$,
$\cos 2 \alpha=\frac{2}{3}, \cos \alpha=\sqrt{\frac{1+\frac{2}{3}}{2}}=\sqrt{\frac{5}{6}}=\frac{\sqrt{30}}{6}$. Therefore $1998 \cos \alpha=333 \quad \sqrt{30}$.