Question 824392
<font color=black size=3>
Let {{{D = sqrt(2)+sqrt(3)}}} and {{{E = sqrt(2)-sqrt(3)}}} so they can be used in a substitution step.


This means 
{{{A = ( sqrt(2)+sqrt(3)+sqrt(5) )( sqrt(2)-sqrt(3)+sqrt(5) )}}}
condenses temporarily to
{{{A = ( D+sqrt(5) )( E+sqrt(5) )}}}


Meanwhile 
{{{B = ( sqrt(2)+sqrt(3)-sqrt(5) )( -sqrt(2)+sqrt(3)+sqrt(5) )}}}
shortens to
{{{B = ( D-sqrt(5) )( -E+sqrt(5) )}}}
or
{{{B = -( D-sqrt(5) )( E-sqrt(5) )}}}
after rearranging a few negative signs.



Then,
{{{AB = 
-(D+sqrt(5))(E+sqrt(5))
(D-sqrt(5))(E-sqrt(5))
}}}
{{{AB = 
-(D+sqrt(5))(D-sqrt(5))
(E+sqrt(5))(E-sqrt(5))
}}}
{{{AB = -(D^2-5)(E^2-5)}}}
Use the difference of squares rule.


For a moment I'll put this main computation on pause.
Let's take a slight detour.


I'll rewrite one of the factors shown above.
I'll start with D and work my way up to D^2-5.
{{{D = sqrt(2)+sqrt(3)}}}
{{{D^2 = (sqrt(2)+sqrt(3))^2}}}
{{{D^2 = (sqrt(2))^2+2*sqrt(2)*sqrt(3)+(sqrt(3))^2}}}
{{{D^2 = 2+2*sqrt(2*3)+3}}}
{{{D^2 = 5+2*sqrt(6)}}}
{{{D^2-5 = 5+2*sqrt(6)-5}}}
{{{D^2-5 = 2*sqrt(6)}}}



Through very similar scratch work, you should have {{{E = sqrt(2)-sqrt(3)}}} become {{{E^2-5 = -2*sqrt(6)}}}



Now returning to the main event.
{{{AB = -(D^2-5)(E^2-5)}}}
{{{AB = -(2*sqrt(6))(-2*sqrt(6))}}}
{{{AB = -2*(-2)*sqrt(6)*sqrt(6)}}}
{{{AB = 4*sqrt(6*6)}}}
{{{AB = 4*sqrt(6^2)}}}
{{{AB = 4*6}}}
{{{AB = 24}}}


I used GeoGebra to verify the answer is correct. 
Another tool you could use is WolframAlpha.
</font>