Can someone help to do this factoring difference of cube?

Question

anonymous · Answer

http://i68.tinypic.com/2chkuac.jpg

anonymous · Answer

16)

welshfella · Answer

use the identity

a^3 - b^3 = (a - b)(a^2 + 2ab + b^2)

anonymous · Answer

I know, but idk this one because when i do i get (a^4-b^4)(a^8+a^4b^4+b^4)

anonymous · Answer

What i need to do next

welshfella · Answer

3a^6 - 3b^6
= 3(a^6 - b^6)
=  3 [ (a^2)^3 - (b^2)^3)]

now use the above identity but replace the a and b  by a^2 and b^2

welshfella · Answer

= 3[(a^2 - b^2)(a^4 + 2a^2b^2 + b^4)]

welshfella · Answer

ok so far?

anonymous · Answer

Sorry but i was asking about another one, that has ^12

anonymous · Answer

Can u help me?? @welshfella

mathmale · Answer

I'd suggest you post only one problem at a time, to eliminate this sort of confusion.

anonymous · Answer

@mathmale i wrote that i need help with 16) . Sorry, I couldnt take different photo

mathmale · Answer

In Problem 16 you have two terms, each of which has the same coefficient (1.2).
Factor out this 1.2.
You can treat the result as a "difference of squares."
The appropriate formula would be a^2 = b^2 = (a+b)(a-b).
Apply this concept to a^12 - b^12.

anonymous · Answer

http://s18.postimg.org/mgyifu4pl/image.jpg Like this?.. @mathmale