sum of integer parts of roots 
If $\alpha,\beta,\gamma$ are real roots of the equation $\displaystyle \frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-2} = \frac{11}{54}x^2.$ Then $\lfloor \alpha \rfloor + \lfloor \beta \rfloor +\lfloor \gamma \rfloor $  

$\bf{Attempt}$ From equation $$\frac{x^2-3x+2+x^2-2x+x^2-x}{x(x-1)(x-2)} = \frac{11}{54}x^2$$
So $$\frac{3x^2-6x+2}{x^3-3x^2+2x} = \frac{11}{54}x^2$$
$$162x^2-324x+108 = 11x^5-33x^4+22x^3$$
$$11x^5-33x^4+22x^3-162x^2+324x-108 =0$$
Could some help me how to solve it, thanks
 A: HINT: use that $$\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x-2}-\frac{11}{54}x^2=-\frac{(x-3) \left(11 x^4+22 x^2-96 x+36\right)}{54 (x-2) (x-1) x}$$
A: If $x\ge 3$, then the LHS is $\le \frac13+\frac12+1=\frac{11}{6}$ and the RHS is $\ge \frac{11}6$ with equality iff $x=3$. Thus we have found our first real root and shown that it is the largest: $x=3$.
If $x<0$, the LHS is negative, the RHS positive, hence we cannot find a real solution there.
If $2<x<3$, then the LHS is strictly decreasing from $+\infty$ to $\frac{11}6$ and the RHS strictly increasing from $\frac{22}{27}$ to $\frac{11}6$. We conclude that there is no real root with floor $2$.
Similarly, in the interval $(1,2)$ and in the interval $(0,1)$, the LHS goes from $+\infty$ to $-\infty$, whereas the RHS remains bounded. We conclude that in each of these intervals, there is at least one real root.
So far, we have found/shown the existence of one real root with $\lfloor x\rfloor =3$, one with $\lfloor x\rfloor =1$, one with $\lfloor x\rfloor =0$. 
If we trust the problem statement that a definite answer can be given at all, we thus see that the answer must be $$4.$$
But for a more satisfying nswer, we have to verify that there are no additional roots in $(0,1)$ and/or $(1,2)$ that would allow us to achieve a different sum with different picks of real roots.
One way you might try for this, is to show that in each of these intervals, the LHS is strictly decreasing.
