If $\lim _{x \rightarrow 0} \dfrac{\left(1+a^3\right)+8 e^{1 / x}}{1+\left(1-b^3\right) e^{1 / x}}=2$, then

Let $L = \lim _{x \rightarrow 0} \dfrac{(1+a^3)+8 e^{1 / x}}{1+(1-b^3) e^{1 / x}}$.

Consider the limit as $x \rightarrow 0^{+}$. As $x \rightarrow 0^{+}$, $1/x \rightarrow \infty$, so $e^{1/x} \rightarrow \infty$.

Dividing the numerator and denominator by $e^{1/x}$, we get:

$L = \lim _{x \rightarrow 0^{+}} \dfrac{(1+a^3)e^{-1/x} + 8}{e^{-1/x} + (1-b^3)} = \dfrac{0 + 8}{0 + (1-b^3)} = \dfrac{8}{1-b^3}$.

Given $L = 2$, we have $\dfrac{8}{1-b^3} = 2$, which implies $1-b^3 = 4$, so $b^3 = -3$, or $b = -3^{1/3}$.

Now consider the limit as $x \rightarrow 0^{-}$. As $x \rightarrow 0^{-}$, $1/x \rightarrow -\infty$, so $e^{1/x} \rightarrow 0$.

$L = \lim _{x \rightarrow 0^{-}} \dfrac{(1+a^3)+8 e^{1 / x}}{1+(1-b^3) e^{1 / x}} = \dfrac{(1+a^3) + 0}{1 + 0} = 1+a^3$.

Given $L = 2$, we have $1+a^3 = 2$, which implies $a^3 = 1$, so $a = 1$.

Thus, $a=1$ and $b=-3^{1/3}$.

Answer: $a=1, b=-3^{1 / 3}$