Tomamos como el límite
$$\lim_{x \to -2^+}\left(\frac{x^{3} \left(x^{3} - 8\right)}{6 x^{2} + 3}\right)$$
cambiamos
$$\lim_{x \to -2^+}\left(\frac{x^{3} \left(x^{3} - 8\right)}{6 x^{2} + 3}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{x^{3} \left(x - 2\right) \left(x^{2} + 2 x + 4\right)}{6 x^{2} + 3}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{x^{3} \left(x^{3} - 8\right)}{3 \left(2 x^{2} + 1\right)}\right) = $$
$$\frac{\left(-2\right)^{3} \left(-8 + \left(-2\right)^{3}\right)}{3 \left(1 + 2 \left(-2\right)^{2}\right)} = $$
= 128/27
Entonces la respuesta definitiva es:
$$\lim_{x \to -2^+}\left(\frac{x^{3} \left(x^{3} - 8\right)}{6 x^{2} + 3}\right) = \frac{128}{27}$$