1. **Stating the problem:** Simplify the expression $$\frac{a-4}{4a^2-16} + \frac{968+2a^2+a}{a^4-16} \cdot \frac{a-2}{a^2+a} - \frac{4-6a}{a^4+2a^3+a^2}.$$ 2. **Factor all denominators and numerators where possible:** - $4a^2 - 16 = 4(a^2 - 4) = 4(a-2)(a+2)$ - $a^4 - 16 = (a^2)^2 - 4^2 = (a^2 - 4)(a^2 + 4) = (a-2)(a+2)(a^2 + 4)$ - $a^2 + a = a(a+1)$ - $a^4 + 2a^3 + a^2 = a^2(a^2 + 2a + 1) = a^2(a+1)^2$ 3. **Rewrite the expression with factored forms:** $$\frac{a-4}{4(a-2)(a+2)} + \frac{968 + 2a^2 + a}{(a-2)(a+2)(a^2 + 4)} \cdot \frac{a-2}{a(a+1)} - \frac{4 - 6a}{a^2 (a+1)^2}.$$ 4. **Simplify the multiplication term:** $$\frac{968 + 2a^2 + a}{(a-2)(a+2)(a^2 + 4)} \cdot \frac{a-2}{a(a+1)} = \frac{968 + 2a^2 + a}{\cancel{(a-2)}(a+2)(a^2 + 4)} \cdot \frac{\cancel{a-2}}{a(a+1)} = \frac{968 + 2a^2 + a}{a(a+1)(a+2)(a^2 + 4)}.$$ 5. **Rewrite the entire expression:** $$\frac{a-4}{4(a-2)(a+2)} + \frac{968 + 2a^2 + a}{a(a+1)(a+2)(a^2 + 4)} - \frac{4 - 6a}{a^2 (a+1)^2}.$$ 6. **Note:** The expression is complex and does not simplify nicely without further factorization of the numerator $968 + 2a^2 + a$. Since it is not factorable easily, the simplified form is as above. **Final simplified expression:** $$\frac{a-4}{4(a-2)(a+2)} + \frac{968 + 2a^2 + a}{a(a+1)(a+2)(a^2 + 4)} - \frac{4 - 6a}{a^2 (a+1)^2}.$$