How do you factor #1 - 256y^8#?

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Answer 1 · 2015-07-15T07:51:29

Use the difference of squares identity three times to find:

#1-256y^8=(1-2y)(1+2y)(1+4y^2)(1+16y^4)#

#1-256y^8#

#=1^2-(16y^4)^2#

#=(1-16y^4)(1+16y^4)#

#=(1^2-(4y^2)^2)(1+16y^4)#

#=(1-4y^2)(1+4y^2)(1+16y^4)#

#=(1^2-(2y)^2)(1+4y^2)(1+16y^4)#

#=(1-2y)(1+2y)(1+4y^2)(1+16y^4)#

...using the difference of squares identity:

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