How do you factor #100a^6-64b^8#?

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Answer 1 · 2015-07-13T00:09:37

#100a^6-64b^8 = 4(5a^3-4b^4)(5a^3+4b^4)#

This is a difference of squares, but first separate out the common scalar factor #4#:

#100a^6-64b^8#

#= 4(25a^6-16b^8)#

#= 4((5a^3)^2-(4b^4)^2)#

#= 4(5a^3-4b^4)(5a^3+4b^4)#

using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with #A = 5a^3# and #B=4b^4#