1. Write the numerator in sigma form

The numerator is

$$(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots+((n-1)^2-(n-1))\cdot 1.$$

This can be written as

$$N_n=\sum_{k=1}^{n-1}(k^2-k)(n-k).$$

Since

$$k^2-k=k(k-1),$$

we expand:

$$N_n=\sum_{k=1}^{n-1}(k^2-k)(n-k) = n\sum_{k=1}^{n-1}(k^2-k)-\sum_{k=1}^{n-1}(k^3-k^2).$$

So,

$$N_n=n\left(\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k\right)-\left(\sum_{k=1}^{n-1}k^3-\sum_{k=1}^{n-1}k^2\right).$$

Thus,

$$N_n=(n+1)\sum_{k=1}^{n-1}k^2-n\sum_{k=1}^{n-1}k-\sum_{k=1}^{n-1}k^3.$$

2. Use standard summation formulas

Let $m=n-1$. Then

$$\sum_{k=1}^{n-1}k=\frac{(n-1)n}{2},$$ $$\sum_{k=1}^{n-1}k^2=\frac{(n-1)n(2n-1)}{6},$$ $$\sum_{k=1}^{n-1}k^3=\left(\frac{(n-1)n}{2}\right)^2.$$

Substitute:

$$N_n=(n+1)\frac{(n-1)n(2n-1)}{6}-n\cdot \frac{(n-1)n}{2}-\left(\frac{(n-1)n}{2}\right)^2.$$

Factor $n(n-1)$:

$$N_n=n(n-1)\left[\frac{(n+1)(2n-1)}{6}-\frac{n}{2}-\frac{n(n-1)}{4}\right].$$

Take LCM $12$ inside the bracket:

$$\frac{2(n+1)(2n-1)-6n-3n(n-1)}{12}.$$

Now,

$$2(n+1)(2n-1)=2(2n^2+n-1)=4n^2+2n-2.$$

So numerator becomes

$$4n^2+2n-2-6n-3n^2+3n=n^2-n-2=(n-2)(n+1).$$

Hence,

$$N_n=\frac{n(n-1)(n-2)(n+1)}{12}.$$

3. Now simplify the denominator

The denominator is

$$D_n=(1^3+2^3+\cdots+n^3)-(1^2+2^2+\cdots+n^2).$$

Using formulas,

$$\sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2,$$ $$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}.$$

Therefore,

$$D_n=\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2n+1)}{6}.$$

Factor $n(n+1)$:

$$D_n=n(n+1)\left[\frac{n(n+1)}{4}-\frac{2n+1}{6}\right].$$

Take LCM $12$:

$$D_n=n(n+1)\cdot \frac{3n(n+1)-2(2n+1)}{12}.$$

Simplify:

$$3n(n+1)-2(2n+1)=3n^2+3n-4n-2=3n^2-n-2=(3n+2)(n-1).$$

So,

$$D_n=\frac{n(n+1)(n-1)(3n+2)}{12}.$$

4. Form the ratio

Thus,

$$\frac{N_n}{D_n} =\frac{\frac{n(n-1)(n-2)(n+1)}{12}}{\frac{n(n+1)(n-1)(3n+2)}{12}} =\frac{n-2}{3n+2}.$$

5. Take the limit

$$\lim_{n\to\infty}\frac{n-2}{3n+2}=\frac{1}{3}.$$

6. Compare with options

So the correct option is

$$\boxed{\frac{1}{3}}.$$

This corresponds to Option D.