If x is positive is x divisible by 2 ? (1) x^3+x is

I don't think that this is a GMAT question. In it's current form it's way beyond the GMAT scope.

First of all note that we are not told that \(x\) is an integer.

Given: \(x>0\). Question: is \(x\) even (integer)?

(1) \(x^3+x=x(x^2+1)=4n\)

If x is an integer then either:
A. \(x\) is a multiple of 4 and \(x^2+1\) is odd;
OR
B. \(x\) is odd and \(x^2+1\) is a multiple of 4. But this scenario is not possible. Why? Because in case \(x=2k+1\) (\(x\) is odd), \(x^2+1=4k^2+4k+2\), which is not a multiple of 4, (it's even never multiple of 4.)
So if \(x\) is an integer it must be multiple of 4.

But \(x\) also can be a non-integer, for example equation \(x^3+x=4\) has one real root (\(x\approx{1.38}\)), which is not an integer OR equation \(x^3+x=8\) also has one real root (\(x\approx{1.83}\)), which is not an integer.

Not sufficient.

(2) \(5x+4=6k\)

If \(x\) is an integer, then \(x\) must be even as \(5x+4=even\) --> \(5x=even-4=even\) --> \(x=even\).

But \(x\) can also be a fraction, for example \(\frac{2}{5}\), or \(\frac{8}{5}\) - basically \(x\) can be a fraction of a form \(x=\frac{a}{5}\), where \(a\) is an integer.

Not sufficient.

(1)+(2) Now if \(x\) is an integer then it's even as we concluded.

But can x be reduced fraction of a form \(x=\frac{a}{5}\)? We'll have \(\frac{a^3}{125}+\frac{a}{5}=4n\) --> \(a^3+25a=a(a^2+25)=500n\) --> as \(a\) and \(n\) are integers, \(a\) must be multiple of 5 (at least) --> hence \(x=\frac{a}{5}=integer\), so \(x\) can not be reduced fraction --> and if \(x\) is an integer, then we know it must be even.

Answer: C.

If we were told that \(x\) is an integer, then the answer would be D.