/sci/ - Science & Math

(1/x^2) = 8 x
1 = 8 x^3
1 = (2 x)^3

oh lord

times both sides by x^2
you get 1=8x^3

then cube root both side
2x=1

>>2392885
up to multiples of the generator of the galois group

thanks for the reply i wasn't thinking that 2^3=8

>>2392885
but 1 has 3 cube roots in the complex numbers, not even to start talking about arbitrary fields.
(Though they of course too can have at most 3 roots of unity.)

>>2392900
lol que?

>>2392908

what you said. x^3=a is of third power and the corresponding galois group has 3 elements. 3 is prime, so its Z/3Z, i.e. 0,1,2 and "+" or another representation would be (1, e^iX, e^2iX), where X is 120°, and " · ".

http://en.wikipedia.org/wiki/Galois_theory

this is why here
>>2392882
I didn't took the root

(1/x^2) = 8x
1 = 8x^3 multiply both sides by x^2
1 = 2x take the third root of both sides (2^3 = 8 and 1^3 = 1)

>>2392919

.....Z/3Z is an infinite quotient ring of integers divisible by 3, so shut the fuck uo about your 1, 2, 3 nonsense sir.

>>2392900
>>2392919
>>2392928
>blockieren

>>2392958
>maximum hurf
the index [Z : 3Z] of the additive subgroup 3Z in Z is finite (it's 3), so all you need is 3 representatives.

Not that you didn't know that already, troll. But just sayin'.

>>2392958
>>2392992
what, where seems to be the problem?
>>2392972
ja!

>>2393098
is she German?

>>2393188
I wish