<s>[INST] If the point of minima of the function ,$f(x)=1+a^2x-x^3$ satisfy the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$, If the point of minima of the function ,$f(x)=1+a^2x-x^3$ satisfy the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$,then $a$ must lie in the interval:
$(A)(-3\sqrt3,3\sqrt3)$
$(B)(-2\sqrt3,-3\sqrt3)$
$(C)(2\sqrt3,3\sqrt3)$
$(D)(-3\sqrt3,-2\sqrt3)\cup (2\sqrt3,3\sqrt3)$

$$f'(x)=a^2-3x^2=0\implies x=\pm\frac{a}{\sqrt3}$$
$$f''(x)=-6x$$ 
So $x=-\frac{a}{\sqrt3}$ is the point of minima.It satisfies the inequality  $\frac{x^2+x+1}{x^2+5x+6}<0$.
$$\frac{x^2+x+1}{x^2+5x+6}<0\implies \frac{1}{x^2+5x+6}<0$$
$$\frac{x^2+5x+6}{(x^2+5x+6)^2}<0\implies x^2+5x+6<0$$
$$(x+2)(x+3)<0\implies (2-\frac{a}{\sqrt3})(3-\frac{a}{\sqrt3})<0$$
$$\implies (\frac{a}{\sqrt3}-2)(\frac{a}{\sqrt3}-3)<0$$
$$a\in (2\sqrt3,3\sqrt3)$$
But the answer is $(-3\sqrt3,-2\sqrt3)\cup (2\sqrt3,3\sqrt3)$
 [/INST] Finding the $x$-value of the minimum point, you have assumed without warrant that $a>0$. We know that the minimum occurs at $x=\pm\frac{a}{\sqrt{3}}$, but we still don't know if $a$ is positive or negative.
Then, you find out that $f''(x)=-6x$, which must be positive at the minimum value. That means we have two cases from here. Either $a>0$ and the minimum occurs at $x=-\frac{a}{\sqrt3}$, or else $a<0$ and the minimum occurs at $x=\frac{a}{\sqrt3}$.
Your reasoning from there is fine; just apply it to both cases.
Backing away from those technical details, we can see that the solution must be symmetric, because we are only given information about $a^2$, so there is no way to tell whether $a$ will be positive or negative. Thus, the solution set for $a$ must be symmetric across the point $a=0$.
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