How do you factor #1331s^12 - 1728t^6#?
1 Answer
#=(sqrt(11)s^2-2sqrt(3)t)(11s^4+2sqrt(33)s^2t+12t^2)(sqrt(11)s^2+2sqrt(3)t)(11s^4-2sqrt(33)s^2t+12t^2)#
Explanation:
We will use the following identities:
-
Difference of squares:
#a^2-b^2 = (a-b)(a+b)# -
Difference of cubes:
#a^3-b^3 = (a-b)(a^2+ab+b^2)# -
Sum of cubes:
#a^3+b^3 = (a+b)(a^2-ab+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^6-b^6#
#= (a^3)^2-(b^3)^2#
#= (a^3-b^3)(a^3+b^3)#
#= (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)#
So we have our very own difference of sixth powers identity:
#a^6-b^6 = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+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#
#(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)#
#=(sqrt(11)s^2-2sqrt(3)t)(11s^4+2sqrt(33)s^2t+12t^2)(sqrt(11)s^2+2sqrt(3)t)(11s^4-2sqrt(33)s^2t+12t^2)#
