How do you factor #1331s^12  1728t^6#?
1 Answer
Answer:
#=(sqrt(11)s^22sqrt(3)t)(11s^4+2sqrt(33)s^2t+12t^2)(sqrt(11)s^2+2sqrt(3)t)(11s^42sqrt(33)s^2t+12t^2)#
Explanation:
We will use the following identities:

Difference of squares:
#a^2b^2 = (ab)(a+b)# 
Difference of cubes:
#a^3b^3 = (ab)(a^2+ab+b^2)# 
Sum of cubes:
#a^3+b^3 = (a+b)(a^2ab+b^2)#
First let us find out how to factorise
We can treat this as a difference of squares, then difference and sum of cubes as follows:
#a^6b^6#
#= (a^3)^2(b^3)^2#
#= (a^3b^3)(a^3+b^3)#
#= (ab)(a^2+ab+b^2)(a+b)(a^2ab+b^2)#
So we have our very own difference of sixth powers identity:
#a^6b^6 = (ab)(a^2+ab+b^2)(a+b)(a^2ab+b^2)#
Note that:

#1331 = 11^3 = (sqrt(11))^6# 
#1728 = 12^3 = (2^2*3)^3 = (2sqrt(3))^6#
Let's put our difference of sixth powers to work with
#a^6  b^6#
#= (sqrt(11)s^2)^6  (2sqrt(3)t)^6#
#= 11^3s^12  12^3t^6#
#= 1331s^12  1728t^6#
#(ab)(a^2+ab+b^2)(a+b)(a^2ab+b^2)#
#=(sqrt(11)s^22sqrt(3)t)(11s^4+2sqrt(33)s^2t+12t^2)(sqrt(11)s^2+2sqrt(3)t)(11s^42sqrt(33)s^2t+12t^2)#