Prove that the value of the expression is negative for any value of a > 1: $$\frac{1}{1-a} + \frac{1}{1+a} + \frac{2}{1+a^2} + \frac{4}{1+a^4} + \frac{8}{1+a^8}$$

Let's simplify the expression:

$$\frac{1}{1-a} + \frac{1}{1+a} + \frac{2}{1+a^2} + \frac{4}{1+a^4} + \frac{8}{1+a^8} = \ \frac{1+a + 1-a}{(1-a)(1+a)} + \frac{2}{1+a^2} + \frac{4}{1+a^4} + \frac{8}{1+a^8} = \ \frac{2}{1-a^2} + \frac{2}{1+a^2} + \frac{4}{1+a^4} + \frac{8}{1+a^8} = \ \frac{2(1+a^2) + 2(1-a^2)}{(1-a^2)(1+a^2)} + \frac{4}{1+a^4} + \frac{8}{1+a^8} = \ \frac{2+2a^2 + 2 - 2a^2}{1-a^4} + \frac{4}{1+a^4} + \frac{8}{1+a^8} = \ \frac{4}{1-a^4} + \frac{4}{1+a^4} + \frac{8}{1+a^8} = \ \frac{4(1+a^4) + 4(1-a^4)}{(1-a^4)(1+a^4)} + \frac{8}{1+a^8} = \ \frac{4+4a^4 + 4 - 4a^4}{1-a^8} + \frac{8}{1+a^8} = \ \frac{8}{1-a^8} + \frac{8}{1+a^8} = \ \frac{8(1+a^8) + 8(1-a^8)}{(1-a^8)(1+a^8)} = \ \frac{8+8a^8 + 8 - 8a^8}{1-a^{16}} = \frac{16}{1-a^{16}}$$

Since (a > 1), (a^{16} > 1), thus (1 - a^{16} < 0). Therefore, the fraction $$\frac{16}{1-a^{16}}$$ is negative.

Answer: The expression is negative for any value of (a > 1).