How do you simplify #(x^3 - 64)/(x^3 + 64) div (x^2 - 16)/(x^2 - 4x + 16)#?

Question

1764 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-14T00:00:23

#(x^3-64)/(x^3+64)-:(x^2-16)/(x^2-4x+16)#

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

#(x^3-64)/(x^3+64)-:(x^2-16)/(x^2-4x+16)#

#=(x^3-4^3)/(x^3+4^3)*(x^2-4x+4^2)/(x^2-4^2)#

#=(cancel(x-4)(x^2+4x+4^2))/((x+4)cancel(x^2-4x+4^2))*(cancel(x^2-4x+4^2))/(cancel(x-4)(x+4))#

#=(x^2+4x+4^2)/(x+4)^2#

#=((x^2+8x+4^2) - 4x)/(x+4)^2#

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

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

with exclusion #x != 4# (coming from the cancelled #(x-4)# term)

..using difference of cubes:

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

..sum of cubes:

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

..and difference of squares:

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