Find the minimum values of $a^3$ and $b^3$ from the following given cubic function. Today after teaching AM,GM,HM relations my teacher proposed this question.
It isn't a HW though, since we don't have to tell that we solved the question and submit a solution. 
Question 

Given that $a,b$ are two positive real numbers.
  If $$f(x)=2x^3+ax^2+bx+4=0$$
  Find the minimum values of $a^3$ and $b^3$.  

Well I have progressed a bit.
Applying $AM\ge GM${taking all the terms in $f(x)$ positive}
We get $8x^6ab\le0$ which is impossible since $x$ can't be $0$ and $a,b$ are positive.
So we have a conclusion that roots of $f(x)$ are negative.
Now putting $x=-X$ where $X$ is positive real number, we get
$2X^3+bX=aX^2+4$ 
This is all useful I got. Using AM,GM on the both the sides one by one yields two inequalities of no use.
Thanks for any solution or hint in advance.

Here's a sketch.
Assume $a, b \geq 0$. Let the roots be denoted $x_1, x_2, x_3$.
Using Vieta's formula we find
\begin{align}
x_1+x_2+x_3 &=  -\frac{a}{2} \\
x_1 x_2 + x_2 x_3 +x_3 x_1&= \frac{b}{2} \\
x1 x_2 x_3 &= -2
\end{align}
The last inequality implies that 
$$2 = (-x_1)(-x_2)(-x_3) = |x_1||x_2||x_3|$$
Applying AM-GM, then
\begin{align}
\frac{a}{6} &= -\frac{1}{3}(x_1+x_2+x_3) \\
&\geq ((-x_1)(-x_2)(-x_3))^{1/3} \\
&= \ldots
\end{align}
Also, 
\begin{align}
\frac{b}{6} &= \frac{1}{3}(x_1 x_2 + x_2 x_3 +x_3 x_1)\\
&\geq (x_1 x_2 x_1 x_3 x_2 x_3)^{1/3}\\
&\vdots
\end{align}
Then, once you have found $a, b$ above the minimums will follow.
