The sum of the series $$\frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \ldots$$ up to 10 terms is

The series to be summed up to 10 terms is $$\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$$ and the general term is given by $$T_n = \frac{n}{1 - 3n^2 + n^4}$$.

Noting that the denominator can be written as $$n^4 - 3n^2 + 1$$, we factor it by treating it as a quadratic in $$n^2$$ to obtain $$(n^2 - n - 1)(n^2 + n - 1)$$. One can verify this factorization by multiplying the factors: $$(n^2 - n - 1)(n^2 + n - 1) = n^4 - 3n^2 + 1$$, which matches the original expression.

With the denominator factored, the general term becomes $$\frac{n}{(n^2 - n - 1)(n^2 + n - 1)}$$. Observing that $$(n^2 + n - 1) - (n^2 - n - 1) = 2n$$ allows us to rewrite the fraction as $$\frac{1}{2} \cdot \frac{(n^2+n-1) - (n^2-n-1)}{(n^2-n-1)(n^2+n-1)} = \frac{1}{2}\left(\frac{1}{n^2 - n - 1} - \frac{1}{n^2 + n - 1}\right)$$.

Defining the function $$f(n) = n^2 - n - 1$$, one finds that $$f(n+1) = (n+1)^2 - (n+1) - 1 = n^2 + n - 1$$. Hence, $$n^2 + n - 1 = f(n+1)$$ and the term simplifies to $$T_n = \frac{1}{2}\left(\frac{1}{f(n)} - \frac{1}{f(n+1)}\right)$$, revealing a telescoping pattern in the series.

When summing from $$n=1$$ to $$10$$, the series telescopes: $$S = \sum_{n=1}^{10} T_n = \frac{1}{2}\sum_{n=1}^{10}\left(\frac{1}{f(n)} - \frac{1}{f(n+1)}\right)$$ reduces to $$\frac{1}{2}\left(\frac{1}{f(1)} - \frac{1}{f(11)}\right)$$ as intermediate terms cancel.

Computing the boundary values gives $$f(1) = 1^2 - 1 - 1 = -1$$ and $$f(11) = 11^2 - 11 - 1 = 109$$. Substituting these into the expression for $$S$$ yields $$S = \frac{1}{2}\left(\frac{1}{-1} - \frac{1}{109}\right) = \frac{1}{2}\left(-1 - \frac{1}{109}\right) = -\frac{55}{109}$$.

Therefore, the sum of the series up to 10 terms is $$-\frac{55}{109}$$.