$\lim _{n \rightarrow \infty}\left[\frac{1}{1-\mathrm{n}^4}+\frac{8}{1-\mathrm{n}^4}+\ldots \ldots \ldots \ldots . .+\frac{\mathrm{n}^3}{1-\mathrm{n}^4}\right]=$

Limit evaluation using cube sum formula.

Combining the terms over the common denominator $1 - n^4$ yields the numerator as the sum of cubes from $1^3$ to $n^3$:

$L = \lim_{n \to \infty} \frac{1^3 + 2^3 + \cdots + n^3}{1 - n^4}$

Applying the known identity $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$:

$L = \lim_{n \to \infty} \frac{n^2(n+1)^2}{4(1 - n^4)} = \lim_{n \to \infty} \frac{n^4 + 2n^3 + n^2}{4 - 4n^4}$

Dividing numerator and denominator by $n^4$:

$L = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{\frac{4}{n^4} - 4}$

As $n \to \infty$, terms with $n$ in the denominator vanish, resulting in:

$L = \frac{1}{-4} = -\frac{1}{4}$

This matches option D.