Question was edited -- please make sure to ask one question per post.

Simplify `((a^6-64))/((a^4+4a^2+16))+((a^4-16))/((a^2+4))` :

(a) `a^6-64` is the difference of two squares `(a^3)^2-(2^3)^2` so it factors as `(a^3-8)(a^3+8)` .

`a^3-8` is the difference of 2 cubes and factors as `(a-2)(a^2+2a+4)`

`a^3+8` is the sum of two cubes and factors as `(a+2)(a^2-2a+4)`

(b) `a^4+4a^2+16` factors as `(a^2+2a+4)(a^2-2a+4)`

(c) `a^4-16` is the difference of two squares and factors as `(a^2+4)(a^2-4)` . Further, `a^2-4` factors as `(a+2)(a-2)`

(d) Finally `a^2+4` is the sum of two squares and will not factor over the reals.

Putting it all together we get:

`(a^6-64)/(a^4+4a^2+16)+(a^4-16)/(a^2+4)=`

`((a-2)(a^2+2a+4)(a+2)(a^2-2a+4))/((a^2+2a+4)(a^2-2a+4))+((a^2+4)(a+2)(a-2))/(a^2+4)`

Cancelling common factors yields:

`=(a-2)(a+2)+(a+2)(a-2)`

`=2(a+2)(a-2)`