i just tried that and got this
(7x-2)^(1/3) + (7x+5)^(1/3) = 3

so i set 7x-2=0 and get x=2/7 ,then i sub 2/7 into original equ


((7(2/7) -2)^(1/3) + (7(2/7) +5)^(1/3) =3
0^(1/3) + (7)^(1/3)=3 then i cube both sides
7=27
and i end up with answer= 27/7

even if it's really simple i'd like to know how the problem is solved lol, i thought it was just something you could solve by substituting and using the cubic formula

wtf r u doin in that post

o i got it i was right but i forgot to do something else

you make it too easy

disregard; im gay

is this rly a troll question or something

and yeah i didn't :~) i thought of something but then i looked at the problem again and it doesn't work

This post was edited by XxAviated on Sep 21 2013 11:25pm

I most likely did something fucking illegal and all mathematicians would like to hang me, but..

(7x-2)^(1/3) + (7x+5)^(1/3) = 3
1/3*log(7x-2) + 1/3*log(7x+5) = log3
log[(7x-2)/3] + log[(7x+5)/3] = log 3
Then I just toss those logs away and get
(7x-2)/3 + (7x+5)/3 = 3
(14x+3)/3 = 3
14x+3=9
14x=6
x=6/14
x=3/7

Which gives the right answer, but I have a very strong feeling I did something I shouldn't have done when I just wiped out the logs...
(And probably when I involved them...)

This post was edited by Dorieus on Sep 22 2013 07:38am

assuming it has to be an easy solution if you are not allowed to use wolfram
let's check if 1+2=3 is the solution
this would need 7x-2=1 and 7x+5=8
the first equation results in x=3/7 and if we plug it into the second hurray it work

so x=3/7 is the solution

Stylin'