How do you factor #8x^3 - 1#?

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Answer 1 · 2015-04-09T03:55:49

Remember this formula for factorizing difference of 2 cubes:
#a^3-b^3 = (a-b)(a^2+ab+b^2)#

In #8x^3-1#,

#a^3=8x^3#
#b^3=1#

#a= root3(8x^3)# = #2x#
#b=root3(1)#= #1#

Substitute #a=2x # , #b=1 # into the formula of #(a-b)(a^2+ab+b^2)#

#(2x-1)##((2x)^2# + (1x#2x#) + #1^2##)# = #(2x-1)##(4x^2# + #2x#+ #1)#

#(2x-1)##(4x^2# + #2x#+ #1)# is the factorized form of #8x^3-1#