Find all possible real number(s) N that satisfy this equation:

⁵√(N³ + 2N) = ³√(N⁵ - 2N)

where, the common value of both sides of the equation is positive.

(In reply to re: Oops! Solution by reading the problem. by Daniel)

Daniel:

Hmm.  Right you are!  I just substituted in N = +/- i, and I see that you are correct.  They are both solutions of the equation.  I wonder where I went wrong.

Also, once again,

  ⁵√(N³ + 2N) = ³√(N⁵ - 2N) = N

I just worked out why.

 If  ³√(N⁵ - 2N) = N
 then (N⁵ - 2N) = N³ by cubing each side
 then  N⁵ = N³ + 2N  by rearranging terms
  and  N = ⁵√(N³ + 2N) by taking the fifth root of each side
  so ⁵√(N³ + 2N) = ³√(N⁵ - 2N) = N
 
I also just noticed that if
If  ³√(N⁵ - 2N) = N
 then (N⁵ - 2N) = N³
 then (N⁵ - N³ - 2N) =   0
 and N ( N^2 + 1)(N^2 - 2) = 0
 and N is 0 or sqrt(2) or -sqrt(2) or i or -i

I am not claiming these are the only solutions to the original equation
     

Posted by Steve Herman on 2009-04-05 23:21:26