各项均为正数的等比数列 $\left\{a_{n}\right\}$ 前 $n$ 项和为 $S_{n}, S_{3}=14, a_{3}=8$, 则 $a_{6}=()$
$: \because \mathrm{S}_{3}=14, \mathrm{a}_{3}=8, \therefore\left\{\begin{array}{l}\frac{a_{1}\left(1-q^{3}\right)}{1-q}=14 \\ a_{1} q^{2}=8\end{array}\right.$, 解得 $\mathrm{a}_{1}=2, \mathrm{q}=2, \therefore \mathrm{a}_{6}=\mathrm{a}_{1} \mathrm{q}^{5}=2 \times 32=64$,

故选:C