How do you factor completely: #49t^6 - 4k^8#?

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Answer 1 · 2015-07-13T13:31:22

This is a difference of squares:

#49t^6-4k^8 = (7t^3)^2-(2k^4)^2 = (7t^3-2k^4)(7t^3+2k^4)#

The difference of squares identity is:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=7t^3# and #b=2k^4# as follows:

#49t^6-4k^8#

#= (7t^3)^2-(2k^4)^2#

#=a^2-b^2#

#= (a-b)(a+b)#

#= (7t^3-2k^4)(7t^3+2k^4)#