4.043. The sum of an infinite geometric progression with a common ratio $|q|<1$ is 4, and the sum of the cubes of its terms is 192. Find the first term and the common ratio of the progression.
## Solution.

From the condition we have

$$
\begin{aligned}
& \left\{\begin{array}{l}
b_{1}+b_{2}+b_{3}+\ldots=4, \\
b_{1}^{3}+b_{2}^{3}+b_{3}^{3}+\ldots=192 \\
|q|<1
\end{array}, \Leftrightarrow\left\{\begin{array}{l}
b_{1}+b_{1} q+b_{1} q^{2}+\ldots=4, \\
b_{1}^{3}+b_{1}^{3} q^{3}+b_{1}^{3} q^{6} \ldots=192,
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ b _ { 1 } ( 1 + q + q ^ { 2 } + \ldots ) = 4 , } \\
{ b _ { 1 } ^ { 3 } ( 1 + q ^ { 3 } + q ^ { 6 } + \ldots ) = 1 9 2 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1} \cdot \frac{1}{1-q}=4, \\
b_{1}^{3} \cdot \frac{1}{1-q^{3}}=192,
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { b _ { 1 } } { 1 - q } = 4 , } \\
{ \frac { b _ { 1 } ^ { 3 } } { ( 1 - q ) ( 1 + q + q ^ { 2 } ) } = 1 9 2 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{b_{1}}{1-q}=4, \\
\frac{b_{1}}{1-q} \cdot \frac{b_{1}^{2}}{1+q+q^{2}}=192
\end{array}\right.\right.
\end{aligned}
$$

$$
\begin{aligned}
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { b _ { 1 } } { 1 - q } = 4 , } \\
{ 4 \cdot \frac { b _ { 1 } ^ { 2 } } { 1 + q + q ^ { 2 } } = 1 9 2 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}=4(1-q), \\
\frac{b_{1}^{2}}{1+q+q^{2}}=48 .
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow \frac{16(1-q)^{2}}{1+q+q^{2}}=48 \Leftrightarrow 2 q^{2}+5 q+2=0, \text { from which } q_{1}=-2, \quad q_{2}=-\frac{1}{2} ;
\end{aligned}
$$

$q_{1}=-2$ does not fit, since $|q|<1$. Then $b_{1}=4\left(1+\frac{1}{2}\right)=6$.

Answer: $6, -\frac{1}{2}$.