Question

S= (a+b)(a²+b²)(a^4+b^4)(a-b)+2b^8
para:
____
a=√√2+1
____
b=√√2-1

Answer 1

$S=(a+b)(a^2+b^2)(a^4+b^4)(a-b)+2b^8\\=(a^2-b^2)(a^2+b^2)(a^4+b^4)+2b^8\\=(a^4-b^4)(a^4+b^4)+2b^8\\=a^8-b^8+2b^8=a^8+b^8\\= (\sqrt{ \sqrt{2}+1 })^8 +(\sqrt{ \sqrt{2}-1 })^8 \\=( \sqrt{2}+1 )^4+( \sqrt{2}-1 )^4\\=(( \sqrt{2}+1 )^2)^2+(( \sqrt{2}-1 )^2)^2\\=(3+2 \sqrt{2} )^2+(3-2 \sqrt{2})^2\\=9+12 \sqrt{2}+8+ 9-12 \sqrt{2}+8\\=34$

S = 34

Saludos!