$r(r-1)^2=1$, how to solve this polynomial analytically? I found this characteristic equation $r(r-1)^2=1$ in my homework Euler differential equation problem (page 669, again the book), $x^{3} y'''+xy'-y=0$, when I misread the problem as
$$x^{3} y'''+xy''-y=0$$
so this beast has characteristic equation that looks nightmarish at least here in WA. I am not sure whether I am missing some smart substitution to proceed, how would you solve the polynomial?
Please, keep the focus on the polynomial. Thanks for pointing out the err but I would like to know how to solve this kind of polynomials. 
 A: You can make the substitution 
$$
z= \displaystyle \frac{1}{r-1}
$$ 
and obtain the equation:
$$z^3 - z - 1 = 0$$
This can be solved cleverly using the substitution $\displaystyle z = \frac{2 \cos \theta}{\sqrt{3}}$ (note: $\displaystyle \theta$ is allowed to be complex) giving rise to the equation
$$ \cos 3 \theta = \frac{3 \sqrt{3}}{2}$$
Thus we get the answer in terms of the complex function $\displaystyle \arccos$ (whose possible arguments range over all the complex numbers)
See here for an explanation of the method: Trigonometric and Hyperbolic Substitution to solve the cubic.
Of course, as Sasha points out, you probably are trying to solve the wrong equation in the first place.
A: Because the ODE you are solving is scale invariant, i.e. if $y(x)$ solves it, so does $y(\lambda x)$ for arbitrary $\lambda \not=0 $, we seek solution as $y(x) = x^r$. This leads to the characteristic equation:
$$
  0 =  r(r-1)(r-2) + r - 1 = r^3 - 3 r^2 + 3r -1 = (r-1)^3
$$
Because the root $r=1$ has triple multiplicity, the solution is logarithmic:
$$
    y(x) = x( c_1 + c_2 \ln(x) + c_3 \ln^2(x) )
$$
A: You could try using the cubic formula.
