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 Fifth root (cubic) = Cube root (quintic) (Posted on 2009-04-05)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution by inspection (spoiler) | Comment 1 of 5
Obviously N = 0 is one solution

For other solutions (N <> 0), raise each side to the 15th power

Then (N^3 + 2N)^3 = (N^5 -2N)^5

N^3 * (N^2 + 2)^3 = N^5 * (N^4 - 2)^5

Dividing by N^3

(N^2 + 2)^3 = N^2 * (N^4 - 2)^5

Let a = N^2

(a + 2)^3 = a*(a^2 -2)^5

a > 0 (because it equals N^2), so the left hand side (lhs) is positive.
therefore the right hand side (rhs) must be positive,
so a^2 must be > 2

when a^2 is > 0, both sides are increasing as a increases.
Clearly, lhs > rhs when a^ = 2, and rhs > lhs when a^2 is a lot bigger, so the rhs gets bigger quickly.  I expect that they will equal at only one point.

I noticed, while writing this, that the point at which they are equal is a = 2.  The lhs = 4^3 = 64.  The rhs = 2*2^5 = 64

so the only other solution is a = 2, which means N = +/- sqrt (2).

Verifying.
If N = sqrt(2),
then N^3 + 2N = 4sqrt(2).  5th root = sqrt(2)
N^3 - 2N = 2sqrt(2).  3rd root = sqrt(2)

If N = -sqrt(2),
then N^3 + 2N = -4sqrt(2).  5th root = -sqrt(2)
N^3 - 2N = -2sqrt(2).  3rd root = -sqrt(2)

Only solutions are N = 0 and N = sqrt(2) and N = -sqrt(2)

Interesting enough, in all cases
5√(N3 + 2N) = 3√(N5 - 2N) = N

Edited on April 5, 2009, 1:20 pm

Edited on April 5, 2009, 1:21 pm
 Posted by Steve Herman on 2009-04-05 13:16:46

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