How do you factor #1331s^12 - 1728t^6#?

1 Answer
George C.
May 15, 2016

#1331s^12 - 1728t^6#

#=(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 #a^6-b^6#

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 = sqrt(11)s^2# and #b=2sqrt(3)t#

#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)#

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