How do you simplify #(1/x + 2/x^2)/(x+8/x^2)#?

1 Answer
George C.
Jan 21, 2017

#(1/x+2/x^2)/(x+8/x^2) = 1/(x^2-2x+4)#

with exclusions #x != 0# and #x != -2#

Explanation:

The sum of cubes identity can be written:

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

So we find:

#(1/x+2/x^2)/(x+8/x^2) = (x^2(1/x+2/x^2))/(x^2(x+8/x^2))" "# with exclusion #x != 0#

#color(white)((1/x+2/x^2)/(x+8/x^2)) = (x^2(1/x+2/x^2))/(x^2(x+8/x^2))#

#color(white)((1/x+2/x^2)/(x+8/x^2)) = (x+2)/(x^3+8)#

#color(white)((1/x+2/x^2)/(x+8/x^2)) = (x+2)/(x^3+2^3)#

#color(white)((1/x+2/x^2)/(x+8/x^2)) = color(red)(cancel(color(black)((x+2))))/(color(red)(cancel(color(black)((x+2))))(x^2-2x+4))#

#color(white)((1/x+2/x^2)/(x+8/x^2)) = 1/(x^2-2x+4)" "# with exclusion #x != -2#

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