Answer
$$\frac{\left(x-2\right)\left(x-8\right)}{x+1}$$
Work Step by Step
Multiplying the two expressions and then cancelling out like terms and factors in the numerator and the denominator, we find:
$$\frac{\left(x^2-10x+16\right)\left(x-1\right)}{x^2-1}\\ \frac{\left(x-2\right)\left(x-8\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\\ \frac{\left(x-2\right)\left(x-8\right)}{x+1}$$
