Factor completely?

#4x^16y^16z^3-4x^16z^3-4y^16z^3+4z^3#

let # x^16 = a#

let #y^16 = b#

let #z^3 =c#

So given expression = #4abc - 4ac - 4bc + 4c#

#=> 4ac ( b - 1) -4 c ( b-1) # ----(group first two and last two terms and take the common factor outside the bracket)

#=> (4ac -4c) (b-1)#

#=> 4c(a-1)(b-1)#

Now substitute the original values of a,b, and c:

#=> 4 z^3 (x^16 -1) (y^16 -1)#

Use identity: #a^2-b^2 = (a-b)(a+b)#

#=> 4 z^3 [(x^8)^2 - 1^2)(y^8)^2 -1^2)] #

#=> 4 z^3 [(x^8 - 1)(x^8 +1)] [(y^8 -1)(y^8 +1)] #

#=> 4 z^3 [(x^8 +1)(y^8 +1)] [(x^8 - 1)(y^8 -1)] #

#=> 4z^3 (x^8+1)(y^8+1) [(x^4)^2 -(1)^2)((y^4)^2 -(1)^2)]#

#=> 4z^3 (x^8+1)(y^8+1) [(x^4+1)(x^4-1)(y^4+1)(y^4-1)]#

#=> 4z^3 (x^8+1)(y^8+1)(x^4+1)(y^4+1)[(x^4-1)(y^4-1)]#

#=> 4z^3 (x^8+1)(y^8+1)(x^4+1)(y^4+1)[(x^2)^2-1^2)(y^2)^2-1^2)]#

#=> 4z^3 (x^8+1)(y^8+1)(x^4+1)(y^4+1)[(x^2+1)(x^2-1)(y^2+1)(y^2-1)]#

#=> 4z^3 (x^8+1)(y^8+1)(x^4+1)(y^4+1)(x^2+1)(y^2+1)[(x^2-1)(y^2-1)]#

#=> 4z^3 (x^8+1)(y^8+1)(x^4+1)(y^4+1)(x^2+1)(y^2+1)[(x-1)(x+1)(y-1)(y+1)]#

#=>4z^3(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)#

#color(red)(4)x^16y^16color(red)(z^3)-color(red)(4)x^16color(red)(z^3)-color(red)(4)y^16color(red)(z^3)+color(red)(4z^3)#,

#=color(red)(4z^3)(x^16y^16-x^16-y^16+1)#,

#=4z^3{x^16(y^16-1)-1(y^16-1)}#,

#=4z^3{(y^16-1)(x^16-1)}#,

#=4z^3[{(y^8)^2-1^2}{(x^8)^2-1^2}#,

#=4z^3{(y^8+1)(y^8-1)}{(x^8+1)(x^8-1)}#,

#=4z^3(x^8+1)(y^8+1){(x^4)^2-1^2}{(y^4)^2-1^2}#,

#=4z^3(x^8+1)(y^8+1){(x^4+1)(x^4-1)}{(y^4+1)(y^4-1)}#,

#=4z^3(x^8+1)(y^8+1)(x^4+1)(y^4+1){(x^2)^2-1^2}{(y^2)^2-1^2}#,

#=4z^3(x^8+1)(y^8+1)(x^4+1)(y^4+1){(x^2+1)(x^2-1)}{(y^2+1)(y^2-1)}#,

#=4z^3(x^8+1)(y^8+1)(x^4+1)(y^4+1)(x^2+1)(y^2+1){(x+1)(x-1)}{(y+1)(y-1)}#,

#=4z^3(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)#.