<s>[INST] 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? 
 [/INST] 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.
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