Answer
$\frac{5x-18}{x^2-2x+4}$
Work Step by Step
Step 1. Factor the third denominator, we have $x^3+8=(x+2)(x^2-2x+4)$
Step 2. The least common denominator can be identified as $(x+2)(x^2-2x+4)$
Step 3. Convert each expression to have the least common denominator then perform operations:
$\frac{5(x^2-2x+4)}{(x+2)(x^2-2x+4)}+\frac{2(x+2)}{(x+2)(x^2-2x+4)}-\frac{60}{(x+2)(x^2-2x+4)}=\frac{5x^2-8x-36}{(x+2)(x^2-2x+4)}=\frac{(x+2)(5x-18)}{(x+2)(x^2-2x+4)}=\frac{5x-18}{x^2-2x+4}$ where $x\ne -2, x^2-2x+4\ne 0$
