For what values of $\alpha > 0$ the equation $p(x) = x^3-9x^2+26x-\alpha =0$ has three positive real roots? 
For what values of $\alpha > 0$ the equation $p(x) = x^3-9x^2+26x-\alpha =0$ has three positive real roots?
The options given are as follows :
$(A)$ $\alpha \ge 27$
$(B)$ $\alpha > 81$
$(C)$ $27 < \alpha <81$
$(D)$ $54 < \alpha \le 81$

What I have tried is that I first transform the equation to the form $y^3 + 3Hy + G = 0$ and then I have applied the fact that this equation has three positive real roots if $G^2 + 4H^3 < 0$.Here I found $G = 24 - \alpha$ and $H = -\frac {1} {3}$ and then calculating we get $\alpha \in (24-\frac {2 {\sqrt 3}} {9} , 24 + \frac {2 {\sqrt 3}} {9})$ which is not similar to any of the given options.Please help me in finding the right option.
Thank you in advance.

Consider the roots are $a$,$b$, $c$.
Consider
$$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$$
So
$$ab+bc+ac=26,\alpha=abc$$
By AM-GM inequality
$$\sqrt[3]{ab\cdot bc\cdot ac}\le\frac{ab+bc+ac}{3}\implies\alpha^{2/3}\le\frac{26}3<9\implies\alpha<27$$
So no options are correct.
