(7x-2)^(1/3) + (7x+5)^(1/3) = 3
(7*12/7 - 2)^(1/3) + (7*12/7 + 5)^(1/3) = 3
(12 - 2)^(1/3) + (12 + 5)^(1/3) = 3
10^(1/3) + 17^(1/3) = 3

considering that 8^(1/3) = 2, we know 8^(1/3) + 8^(1/3) = 4 which is greater than 3, but somehow 10^(1/3) + 17^(1/3) is equal to 3?

but what do i know. you're right, the answer is 12/7. we should smack saber for giving such an easy problem that two people here already got faster than OP could respond.

This post was edited by carteblanche on Sep 21 2013 08:49pm

somehow plugging the equation back in, without cubing each side again, it fails. However, if you cube each side during plugging in, it works. I don't quite understand why, but there you have it lol.

Agreed, it was a simple equation, perhaps he's just learning Algebra.

it's because of the distributive property.

(a + b)^2 = (a+b)(a+b) != a^2 + b^2

Yeah... don't forget to distribute guys. But whatever, I think this is a job for Wolfram

you can't cube each individual term. You have to cube the whole expression on each side if you do that.

Figured something like that.

When I solve it without cubing each side until the very end, I get x=42.875 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; x^(1/3) = 2/7 <--- That could work if you keep x^(1/3), haven't tested it though lol.
Which is definitely not right haha

Anyway, im just using this to procrastinate. I should get back to my paper haha.

This post was edited by smoothy9 on Sep 21 2013 09:19pm

lmao glanced at it the first time but i stared at it for a minute the second time before deciding i didn't do anything wrong. can't believe i did that

yes sorry lul, help me please though
kdawg help

This post was edited by saber_x3 on Sep 21 2013 10:38pm

you can probably use substitution to solve then but im still talking out of my ass

i liked the other problem someone posted a while back. 2^x = 1/x