why ?

just expand it brainlet

Why not?

>>8777733

can't argue with these digits

reeeeeeeeee

>>8777726
1. expand them
2. turn the results into mixed radicals (if possible)
3. add or subtract the the radicals with the same sqrt number

>>8777790

too long and not interesting

isn't there a shortcut ?

>>8777811
[eqn](a+b)(a-b)=a^2-b^2[/eqn]
Rinse, repeat

File: Screenshot from 2017-03-25 17-12-06.png (31KB, 889x304px) Image search: [Google]

31KB, 889x304px

>>8777726
Because all of the radicals cancel out regardless of what numbers you use.

[eqn](\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})(\sqrt{a}-\sqrt{b}+\sqrt{c}+\sqrt{d})\\ (\sqrt{a}+\sqrt{b}-\sqrt{c}+\sqrt{d})(\sqrt{a}+\sqrt{b}+\sqrt{c}-\sqrt{d})\\ (\sqrt{a}-\sqrt{b}-\sqrt{c}+\sqrt{d})(\sqrt{a}-\sqrt{b}+\sqrt{c}-\sqrt{d})\\ (\sqrt{a}+\sqrt{b}-\sqrt{c}-\sqrt{d})(\sqrt{a}-\sqrt{b}-\sqrt{c}-\sqrt{d})\\ =((a-b)^2+(c-d)^2-2(a+b)(c+d))^2-64abcd[/eqn]

>>8777777

>>8777726
Notice how there is an even number of bracket terms (8) so if you take any one of the radicals x^(1/2) and its equivalent in another bracket it will always give x^(8/2)=x^4
Notice also how between any two radicals, 4 of the bracket terms have a plus sign and 4 have a minus sign. This probably plays into the law (1-x)(1+x)=1-x^2 in some way. Something like that.

interesting
that's the first 3 primes, and all permutations of positive and negative roots of them

and it ends up being -71, also a prime