Question

find the value

Answer 1

42?

Answer 2

find the value of the expression\[\frac{a^{32} -b^{32}}{(a^2 + b^2)(a^4 + b^4)(a^8 + b^8)(a^{16} + b^{16})} + 12b\] if the only thing known about a and b is that
a + b = 6

Answer 3

eww

Answer 4

36

Answer 5

\[a+b=6 \\ (a+b)^2=36 \\a^2+2ab+b^2=36 \\ a^2+b^2=36-2ab \\ (a+b)^4=1296 \\ a^4+4a^3b+6a^2b^2+4ab^3+b^4=1296 \\ a^4+b^4=1296-4a^3b-6a^2b^2-4ab^3\]
blah blah
this seems like an awful long way

Answer 6

maybe we can factor the top and see if something cancels

Answer 7

@Zarkon you need a group for yourself above champion, perhaps demigod

Answer 8

the numerator factors nicely

Answer 9

difference of squares (over and over)

Answer 10

\[a^{32}-b^{32}=(a^{16}-b^{16})(a^{16}+b^{16}) \\ =(a^4-b^4)(a^4+b^4)(a^{16}+b^{16})\]
little annoying using that over and over

Answer 11

but prettier

Answer 12

not 4... it is 8

Answer 13

lol yeah i was testing you guys

Answer 14

\[a^{32}-b^{32}=(a^{16}-b^{16})(a^{16}+b^{16}) \\ =(a^8-b^8)(a^8+b^8)(a^{16}+b^{16}) \]:p

Answer 15

and then so on...

Answer 16

\[a^{32} -b^{32}=(a-b)(a+b)(a^2 + b^2)(a^4 + b^4)(a^8 + b^8)(a^{16} + b^{16})\]

Answer 17

\[(a-b)(a+b)+12b \\(a-b)6+12b \\ 6a-6b+12b \\ 6a+6b =6(a+b)\]

Answer 18

it is weird your 42 wasn't to far off

Answer 19

nope ;)