1. **State the problem:** Simplify the expression $$\frac{(n-1)(n+1)^{-2}-2n(n^2-1)^{-1}+(n+1)(n-1)^{-2}}{8n(n^4-1)^{-1}}$$ 2. **Recall important formulas and rules:** - Recall that $n^2-1 = (n-1)(n+1)$. - Negative exponents mean reciprocal: $x^{-m} = \frac{1}{x^m}$. - When simplifying complex fractions, multiply numerator and denominator by common denominators to clear fractions. 3. **Rewrite terms with positive exponents:** - $(n+1)^{-2} = \frac{1}{(n+1)^2}$ - $(n^2-1)^{-1} = \frac{1}{n^2-1} = \frac{1}{(n-1)(n+1)}$ - $(n-1)^{-2} = \frac{1}{(n-1)^2}$ - $(n^4-1)^{-1} = \frac{1}{n^4-1}$ 4. **Rewrite numerator:** $$ (n-1)\frac{1}{(n+1)^2} - 2n \frac{1}{(n-1)(n+1)} + (n+1) \frac{1}{(n-1)^2} = \frac{n-1}{(n+1)^2} - \frac{2n}{(n-1)(n+1)} + \frac{n+1}{(n-1)^2} $$ 5. **Find common denominator for numerator terms:** The common denominator is $(n-1)^2 (n+1)^2$. Rewrite each term: - $\frac{n-1}{(n+1)^2} = \frac{(n-1)(n-1)^2}{(n+1)^2 (n-1)^2} = \frac{(n-1)^3}{(n-1)^2 (n+1)^2}$ - $\frac{2n}{(n-1)(n+1)} = \frac{2n (n-1)(n+1)}{(n-1)^2 (n+1)^2} = \frac{2n (n-1)(n+1)}{(n-1)^2 (n+1)^2}$ - $\frac{n+1}{(n-1)^2} = \frac{(n+1)(n+1)^2}{(n-1)^2 (n+1)^2} = \frac{(n+1)^3}{(n-1)^2 (n+1)^2}$ So numerator becomes: $$ \frac{(n-1)^3 - 2n (n-1)(n+1) + (n+1)^3}{(n-1)^2 (n+1)^2} $$ 6. **Simplify numerator's numerator:** Expand each term: - $(n-1)^3 = n^3 - 3n^2 + 3n -1$ - $-2n (n-1)(n+1) = -2n (n^2 -1) = -2n^3 + 2n$ - $(n+1)^3 = n^3 + 3n^2 + 3n +1$ Sum: $$ (n^3 - 3n^2 + 3n -1) + (-2n^3 + 2n) + (n^3 + 3n^2 + 3n +1) = $$ $$ (n^3 - 2n^3 + n^3) + (-3n^2 + 3n^2) + (3n + 2n + 3n) + (-1 + 1) = $$ $$ (0) + (0) + (8n) + (0) = 8n $$ 7. **So numerator simplifies to:** $$\frac{8n}{(n-1)^2 (n+1)^2}$$ 8. **Rewrite denominator:** $$8n (n^4 -1)^{-1} = 8n \frac{1}{n^4 -1}$$ Recall $n^4 -1 = (n^2 -1)(n^2 +1) = (n-1)(n+1)(n^2 +1)$ 9. **Rewrite denominator as:** $$\frac{8n}{(n-1)(n+1)(n^2 +1)}$$ 10. **The entire expression is:** $$ \frac{\frac{8n}{(n-1)^2 (n+1)^2}}{\frac{8n}{(n-1)(n+1)(n^2 +1)}} = \frac{8n}{(n-1)^2 (n+1)^2} \times \frac{(n-1)(n+1)(n^2 +1)}{8n} $$ 11. **Cancel common factors:** $$ = \frac{\cancel{8n}}{(n-1)^2 (n+1)^2} \times \frac{(n-1)(n+1)(n^2 +1)}{\cancel{8n}} = \frac{(n-1)(n+1)(n^2 +1)}{(n-1)^2 (n+1)^2} $$ 12. **Cancel common factors:** $$ = \frac{\cancel{(n-1)} \cancel{(n+1)} (n^2 +1)}{(n-1) \cancel{(n-1)} (n+1) \cancel{(n+1)}} = \frac{n^2 +1}{(n-1)(n+1)} $$ 13. **Recall denominator:** $$(n-1)(n+1) = n^2 -1$$ 14. **Final simplified expression:** $$\boxed{\frac{n^2 +1}{n^2 -1}}$$ This is the simplified form of the given expression.