How do you factor #(2x+1)^3 + (2y)^3#?

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Answer 1 · 2015-05-28T19:46:23

For any numbers a and b, #a^3 + b^3 = (a+b)(a^2-ab+b^2)#

So, this expression becomes:

#((2x+1)+2y)((2x+1)^2-(2x+1)(2y)+(2y)^2)#

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

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

These 2 numbers look like they can be factored, but they can't because x and y are different variables.

Answer 2 · 2015-05-28T19:48:13

You can use the identity for a sum of cubes:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

substituting #a=2x+1# and #b=2y# to get:

#(2x+1)^3+(2y)^3#

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

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

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