$\begin{aligned}\begin{cases}u_1+u_2+u_3=14\\ u_1.u_2.u_3=64\end{cases}&\Leftrightarrow\begin{cases}u_1+u_1.q+u_1.q^2=14\\ u_1.u_1.q.u_1.q^2=64
\end{cases}\\ &\Leftrightarrow\begin{cases}u_1+u_1.q+u_1.q^2=14\\ \left(u_1.q\right)^3=4^3\end{cases}\\ &\Leftrightarrow\begin{cases}u_1+4+u_1.q^2=14\\ u_1.q=4\end{cases}\\ &\Leftrightarrow\begin{cases}u_1+u_1.q^2=10\\ u_1.q=4\end{cases}\\ &\Leftrightarrow\begin{cases}u_1\left(1+q^2\right)=10 &(1)\\ u_1.q=4 &(2)\end{cases}\end{aligned}$

Chia (1) cho (2) ta được: $$\begin{aligned}\dfrac{1+q^2}{q}=\dfrac{10}{4}=\dfrac{5}{2}&\Leftrightarrow2q^2-5q+2=0\\ &\Leftrightarrow\left[\begin{array}{l}q=2\\ q=\dfrac{1}{2}\end{array}\right.\\ &\overset{(2)}{\mathop{\Leftrightarrow }}\left[\begin{array}{l}u_1=\dfrac{4}{q}=2\\
u_1=\dfrac{4}{q}=8\end{array}\right.\end{aligned}$$