Does that sound rational to you?

Since this polynomial can be written as a polynomial in $nx$ with integer coeffiecients, any rational solution to this equation can be written as $x=c/n$ for some integer $c$ , where $\begin{array}{rcl} nc^3 - n(n+2)c^2 - n^3c + n^3(n-8) & = & 0 \\ c^3 - (n+2)c^2 - n^2c + n^2(n-8) & = & 0 \\ c^2(c-n-2) & = & n^2(c - n + 8) \\ c^2d & = & n^2(d+10) \end{array}$ so that $\begin{array}{rcl} 10n^2 & = & d(c^2-n^2) \; = \; d(c-n)(c+n) \\ & = & d(d+2)(d+2+2n) \end{array}$ where $d = c-n-2$ . Thus $\begin{array} {rcl} 10n^2 - 2d(d+2)n & = & d(d+2)^2 \\ \big[10n - d(d+2)\big]^2 & = & d^2(d+2)^2 + 10d(d+2)^2 \\ & = & d(d+10)(d+2)^2 \end{array}$ so we deduce that $d(d+10) = e^2$ must be a perfect square. This implies that $(d+5)^2 = e^2 + 25$ , and hence that $(d+5+e)(d+5-e) = 25$ . This equation enables us to determine all possible values of $d,e$ . Since $n = \tfrac{1}{10}(d+2)(d\pm e)$ must be an integer, we can deduce the following possibilities: $\begin{array}{|c|c|c|c|c|c|} \hline d+5+e & d+5-e & d & e & n & c \\ \hline 25 & 1 & 8 & 12 & -4,20 & 6,30 \\ \hline 5 & 5 & 0 & 0 & 0 & 2 \\ \hline 1 & 25 & 8 & -12 & -4,20 & 6,30 \\ \hline -25 & -1 & -18 & -12 & 48 & 32\\ \hline -5 & -5 & -10 & 0 & 8 & 0\\ \hline -1 & -25 & -18 & 12 & 48 & 32\\ \hline \end{array}$ Thus we obtain possible values of $n$ being $-4,20,48,8$ , with corresponding rational roots $-\tfrac32,\tfrac32,\tfrac23,0$ . Thus the required answer is $-4+20+48+8 = 72$ .

Separate out to get $n(x^3-x^2-x+1) = 2x^2+8.$ Write $x = p/q$ , with $p$ and $q$ coprime, multiply by $q^3$ , and factor the left side to get $n(p-q)^2(p+q) = q(2p^2+8q^2).$ Take the equation mod $p-q$ to get $0 \equiv 10q^3$ mod $p-q$ . Take the equation mod $p+q$ to get $0 \equiv 10q^3$ mod $p+q$ . Since $q$ and $p \pm q$ are relatively prime, we get that $p-q$ and $p+q$ both divide $10$ .

There are eight divisors of $10$ to choose from for $p-q$ and $p+q$ . Also they have to be congruent mod $2$ and $p-q < p+q$ . That leads to the following possibilities for $(p-q,p+q)$ : $(-10,-2), (-10,2), (-10,10), (-5,-1), (-5,1),$ $(-5,5), (-2,2), (-2,10), (-1,1), (-1,5), (1,5), (2,10)$ All of the pairs of even numbers happen to lead to values of $p$ and $q$ that are both even, which is no good; so that leaves $(-5,-1), (-5,1), (-1,1), (-1,5), (1,5).$ This leads to $p/q = -3/2, -2/3, 0, 2/3, 3/2$ respectively. Plugging in, setting to $0$ , and solving for $n$ , we get $-4, 9.6, 8, 48, 20$ respectively. Throwing out the non-integer value and taking the sum, we get $\fbox{72}$ .

Add 10 to both sides and factor as (x^2-1)(n(x-1)-2)=10. Let y=x-1, then y(y+2)(ny-2)=10. Let y=p/q, relatively prime of course. Then p(p+2q)(np-2q)=10q^3. Since they are relatively prime p|10 and WLOG p is positive, we can let q be negative if we want.

Case 1. p=1. (1+2q)(n-2q)=10q^3. Note that 8q^3-(4q^2-2q+1)(1+2q)=-1 so q^3 and 1+2q are relatively prime; 1+2q must divide 10. Then q=-3, -1, 0, 2 which yield n=48, 8, and 20.

Case 2. p=2. 8(1+q)(n-q)=10q^3. Then q must be even, but p is even so they cant be relatively prime. We already covered that case.

Case 3. p=5. (5+2q)(5n-2q)=2q^3. Note that 8q^3-(4q^2-10q+25)(2q+5)=-125 so 2q+5 divides 125: q=-65, -15, -5, -3, -2, 0, 10, 60. Testing these yields that n=48, 8, -4, or 20. Some will yield noninteger values of n.

So thus n=48, 8, -4, or 20. Their sum is 72.