GOOD QUESTION

let p be the only real root of the equation.

then we have p³-(a-4)(a+2)p - a = 0 -------[1ST]

now, (x-p) is a factor of x³ - (a-4)(a+2)x - a

we can write,
x³ - (a-4)(a+2)x - a

=(x-p)[x² + px +p² - (a-4)(a+2)]

Now, as p is the only root ,[x² + px +p² - (a-4)(a+2)] has discriminant less than zero

so, p² - 4{p² - (a-4)(a+2)} < 0

or, 4(a-4)(a+2) - 3p² < 0

now if the above equation is true for min value of 3p² (which is zero), then it is true for all real p

so 4(a-4)(a+2)<0

which gives, a lies between (-2) and 4

That's right. here's another take on the problem: x^3 -(a^2-2a-8)x -a = 0

has only one root implies that its derivative has at most one root.

i.e. g(x) = 3x^2 -(a^2-2a-8) has at most one real root. If a=4 or a=-2, then x=0 is the only real root. If (a^2-2a-8)>0 then we have two distinct roots. Hence (a^2-2a-8)<0

This is also sufficient as if (a^2-2a-8)<0, then the given cubic is monotonic and therefore has exactly one root.

Hence a \in [-2,4]