(x2-1)/(√(x5+x4+x3)+√(x5-x4+x3))=
(x2-1)/[x(√(x3+x2+x)+√(x3-x2+x)]=
[(x2-1)/x]((√(x3+x2+x)-√(x3-x2+x))/(x3+x2+x-x3+x2-x) (after rationalisation by multiplying top and bottom by √(x3+x2+x)-√(x3-x2+x))=
(x+1)(x-1)((√(x3+x2+x)-√(x3-x2+x))/(2x3).
x4-x=x(x3-1)=x(x-1)(x2+x+1); x4+x=x(x+1)(x2-x+1).
Therefore x3+x2+x=(x4-x)/(x-1); x3-x2+x=(x4+x)/(x+1).
(x+1)(x-1)((√(x3+x2+x)-√(x3-x2+x))/(2x3)=
(x+1)(x-1)((√((x4-x)/(x-1))-√((x4+x)/(x+1)))/(2x3)=
(x+1)√(x-1)√(x4-x)-(x-1)√(x+1)√(x4+x)/(2x3)=
(x+1)√((x-1)(x4-x))-(x-1)√((x+1)(x4+x))/(2x3)=
(x+1)√(x5-x4-x2+x)-(x-1)√(x5+x4+x2+x)/(2x3).
The expressions shown in green (including the given one) are equivalent.
The expressions in other colours are equivalent (in the same colour), but are different to expressions in different colours, even though they appear to be derived from one another.
The reason is based on behaviour for 0<x<1, when √(x-1) does not exist and for x=1, when some quantities are undefined. When x=1, all the expressions=0. When x=0 the green expressions are undefined but as x→0, the expressions→-∞. The blue expressions→∞ (positive) as x→0. The red expression cannot be evaluated because of √-1. It is ambiguous whether the expression tends to either positive or negative infinity.
More to follow...