There are precisely two nonnegative real values of R, so that for each of these values:

³√(3 + √R) + ³√(3 - √R) is an integer.

Find these values, and prove that no other nonnegative real R can conform to the given conditions.

*** While the solution may be facile with the aid of a computer program, show how to derive it without one.

as with xdogs solution let A=3+sqrt(r), B=3-sqrt(r) and I=A^(1/3)+B^(1/3)

(1)  I^3=A+3(A^2B)^(1/3)+3(AB^2)+B

now A+B=6 so

I^3-6=3[ (A^2B)^1/3 + (AB^2)^1/3 ]

C=(I^3-6)/3=(A^2B)^1/3 + (AB^2)^1/3

C^3=A^2B+3AB(A^2B)^1/3+3AB(AB^2)^1/3+AB^2

C^3=AB(A+3(A^2B)^1/3+3(AB^2)^1/3+B)

now we can use (1) to substitute I^3 for the part in the parenthesis

C^3=ABI^3

(I^3-6)^3/27=ABI^3

AB=9-R

9-R=(I^3-6)^3/(27I^3)

(2) R=9-(I^3-6)^3/(27I^3)

and this agrees with my original solution

a further note (2) can be further generalized if be substitute x for the two 3's in the main equation then (2) becomes

R=x^2-(I^3-2x)^3/(27x^3)

Posted by Daniel on 2008-08-22 22:00:37