Let us first expand #{1+2x}/{1-2x}# in powers of #x# up to the third power (since we want the final expression only up to the third power we don't need to worry about the subsequent terms - they would always contribute higher powers.
For this, we use the result
#1/{1-x} = 1+x+x^2+x^3+x^4 +...#
and get
#{1+2x}/{1-2x} = (1+2x)(1+2x+4x^2+8x^3+...)#
# =1 times (1+2x+4x^2+8x^3 + ... )#
#+2x times (1+2x+4x^2+8x^3 + ... )#
#= (1+2x+4x^2+8x^3 + ... )+ (2x+4x^2+8x^3 + ... )#
Where we have dropped #2x times 8x^3# in the second bracket above since it goes beyond #x^3#
So
#{1+2x}/{1-2x} = 1+4x+8x^2+16x^3+...#
We next use the binomial expansion for
#(1+x)^{1/2} = 1+1/2 x+{1/2(-1/2)}/{2!}x^2+{1/2(-1/2)(-3/2)}/{3!}x^3+...= 1+1/2 x-1/8 x^2+1/16 x^3 +...#
Here again we drop the higher order terms, because they will not contribute in the final result up to #x^3#.
Thus
#sqrt{{1+2x}/{1-2x}} = [1+(4x+8x^2+16x^2+..)]^{1/2}.#
#= 1 + 1/2 (4x+8x^2+16x^3+..)#
#-1/8 (4x+8x^2+16x^3+..)^2 +1/16 (4x+8x^2+16x^3+..)^3+..#.
#= 1+(2x+4x^2+8x^3 + ...)-1/8(16x^2+2times 4x times 8x^2+... )+1/16 times (64x^3+...)+ldots#
#= 1+2x+(4-2)x^2+(8-8+4)x^3+...#
#=1+2x+2x^2+4x^3+...#
