Thought I would Close Things Up with this Algebra Problem.

This came to my attention that a User by the name of StilTzkin.

"My girlfriend's math teacher was offering this problem for extra credit:
(7x-2)^(1/3) + (7x+5)^(1/3) = 3
It was due today so it's too late to get extra credit for it, but it's been driving me and my friend completely insane because we are usually pretty proud of our math skills, but this one has us completely stumped. We know (thanks to an online math solver) that the solution is 3/7, but the point was to show steps. We tried using logs but that led us nowhere, and cubing the whole thing leads us to a dead end as well. Can someone at least give me a hint or something? thanks"

I thought I would give some very easy Algebraic help to solve this extremely easy problem

We start by realizing what this is (7x - 2) ^(1/3) is the same as the cube root of 7x-2, which means cubing it will get rid of the 1/3 exponent.

(7x-2)^(1/3) + (7x+5)^(1/3) = 3
We can get rid of the 1/3 powers just by cubing both sides, cubing both sides will give us

7x-2 + 7x+5 = 27
Now we have a very simple algebraic expression

14x + 3 = 27
14x = 24
x = 12/7

There was some discussion on whether this was the right answer, but it actually is intact correct, lets use the same methods we did to solve it, to check it.

(7[12/7] - 2)^(1/3) + (7[12/7] + 5)^(1/3) = 3

The 7s will cancel leaving us with

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

Lets cube both sides like we did earlier

10 + 17 = 27
27 = 27
It checks out.

 

What? What the?! WTF!

 

Algebra? I can't hear you la, la, la, la, la, la, I can't hear you, la, la, la, la, la ,la.