4.1. (13 points) The numerator of an irreducible fraction was cubed, and the denominator was increased by 3. As a result, the fraction doubled. Find the value of the original fraction, if its numerator and denominator are natural numbers.
Answer: $\frac{2}{3} \approx 0.67$.

Solution. According to the condition $2 \cdot \frac{a}{b}=\frac{a^{3}}{b+3}$, from which $a^{2}=\frac{2(b+3)}{b}=2+\frac{6}{b}$. Therefore, 6 is divisible by $b$. If $b=1$, 2 or 6, then $a^{2}=8,5$ or 3 respectively. Such integers $a$ do not exist. If $b=3$, then $a=2$. The required fraction: $\frac{2}{3}$.