It Is A Polynomial, Isn't it?

The equation can be written

$\sum_{x=1}^n x^{3} - \sum_{x=1}^n 46x + \sum_{x=1}^n \frac{360}{n} = 0$

$\sum_{x=1}^n x^{3} - 46 \times \sum_{x=1}^n x + n \times \frac{360}{n} = 0$

$(1^{3} + 2^{3} + 3^{3} + ... + n^{3}) - 46(1 + 2 + 3 + ... + n) + 360 = 0$

We have

$(1^{3} + 2^{3} + 3^{3} + ... + n^{3}) = (1 + 2 + 3 + ... n)^{2}$

and

$1 + 2 + 3 + ... n = \frac{(n)(n+1)}{2}$

So the equation become

$(\frac{(n)(n+1)}{2})^{2} -46(\frac{(n)(n+1)}{2}) + 360 = 0$

$(\frac{(n)(n+1)}{2} - 10)(\frac{(n)(n+1)}{2} - 36)$

$\frac{(n)(n+1)}{2} = 10$ or $\frac{(n)(n+1)}{2} = 36$

for $\frac{(n)(n+1)}{2} = 10$

$n^{2} + n = 20$

$n^{2} + n - 20 = 0$

$(n + 5)(n - 4) = 0$

$n = -5$ doesn't satisfy $n > 1$ so $n = 4$

for $\frac{(n)(n+1)}{2} = 36$

$n^{2} + n = 36$

$n^{2} + n - 36 = 0$

$(n + 9)(n - 8) = 0$

$n = -9$ doesn't satisfy $n > 1$ so $n = 8$

So we get the sum of possible values of $n$ is $4 + 8 = \boxed{12}$

*cmiiw