Condition for $f(x)=2x^3+ax^2+bx+4$ to have three real roots If $2x^3+ax^2+bx+4=0$ ( $a$ and $b$ are positive real numbers). Find minimum value of $a^3$ and $b^3$ so that the equation has $3$ distinct real roots.
Now $f(x)=2x^3+ax^2+bx+4$
For three distinct roots,
$f'(x)=6x^2+2ax+b$ must have two distinct real roots. Let us say roots are $p$ and $q$. Now for three real roots, should the condition $f(p) \times f(q)<0$ also be satisfied because if $f(p)$ and $f(q)$ have same sign, then we will get only one real root.
Is my approach correct? And is there any better way to approach this? 

Your method is fine so far, but I think it's worth pointing out a way to make the subsequent calculation easier. Since $p$ is a root of $6x^2+2ax+b=0$, we have
$$ p^2 = \frac{-2ap-b}{6}, $$
which simplifies the calculation of $f(p)$ to a linear one:
\begin{align}
f(p) &= 2p^3 + ap^2 + bp+4 \\
&= (2p+a)\frac{-2ap-b}{6} +bp+4 \\
&= -\frac{2a}{3}p^2-\frac{a^2+b}{3}p -\frac{ab}{6}+bp+4 \\
&= \frac{2a}{3} \frac{2ap+b}{6} +\frac{1}{3}(2b-a^2) p + 4-\frac{ab}{6} \\
&= \frac{6b-a^2}{9}p + 4-\frac{ab}{18}
\end{align}
Of course, exactly the same thing holds for $q$. But now we can expand the product and use the formulae
$$ p+q = -\frac{a}{3}, \qquad pq = \frac{b}{6} $$
to remove $p$ and $q$ entirely:
\begin{align}
f(p)f(q) &= \left(\frac{6b-a^2}{9}p + 4-\frac{ab}{18} \right)\left(\frac{6b-a^2}{9}q + 4-\frac{ab}{18} \right) \\
&= \frac{(6b-a^2)^2}{81}pq + \frac{12b-a^2}{9} \left(4-\frac{ab}{18}\right)(p+q) + \left( 4-\frac{ab}{18} \right)^2 \\
&= \dotsb = \frac{1}{108}(1728+16a^3+8b^3-a^2b^2-144ab)
\end{align}
We want this to be nonpositive. Also, from the discriminant of the quadratic, we have $a^2-6b \geq 0$. If we suppose that $a^2=6b$, the equation we obtained from the cubic becomes
$$ 1728 -8a^3 + \frac{1}{108}a^6, $$
a quadratic in $a^3$ with one real root, a double root at $a^3=432$. Hence this is the only possible answer, and plotting the regions where the inequalities are satisfied confirms this. $b^3>864$ is the other inequality.
