The ratio of the coefficient of $x^3$ to the term independent of $x$ in the expansion of $\left(2 x+\dfrac{1}{x^2}\right)^{12}$ is

The general term in the expansion of $\left(2x + \dfrac{1}{x^2}\right)^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2x)^{12-r} \left(\dfrac{1}{x^2}\right)^{r} = ^{12}C_{r} 2^{12-r} x^{12-r} x^{-2r} = ^{12}C_{r} 2^{12-r} x^{12-3r}$.

To find the coefficient of $x^3$, set the exponent $12 - 3r = 3$, which gives $3r = 9$, so $r = 3$.

The coefficient of $x^3$ is $^{12}C_{3} 2^{12-3} = ^{12}C_{3} 2^{9} = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1} \times 2^9 = 220 \times 512$.

To find the term independent of $x$, set the exponent $12 - 3r = 0$, which gives $3r = 12$, so $r = 4$.

The term independent of $x$ is $^{12}C_{4} 2^{12-4} = ^{12}C_{4} 2^{8} = \dfrac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times 2^8 = 495 \times 256$.

The ratio of the coefficient of $x^3$ to the term independent of $x$ is $\dfrac{220 \times 512}{495 \times 256} = \dfrac{220}{495} \times 2 = \dfrac{44}{99} \times 2 = \dfrac{4}{9} \times 2 = \dfrac{8}{9}$.

Answer: $8: 9$