Analizing the stability of the equilibrium points of the system $\ddot{x}=(x-a)(x^2-a)$ $\require{amsmath}$
$\DeclareMathOperator{\Tr}{Tr}$
$\DeclareMathOperator{\Det}{Det}$

Investigate the stability of the equilibrium points of the system $\ddot{x}=(x-a)(x^2-a)$ for all real values of the parameter $a$. (Hints: It might help to
  graph the right-hand side. An alternative is to rewrite the equation as $\ddot{x}=−V′(x)$ for a suitable potential energy function $V$ and then use your intuition about particles
  moving in potentials.)

I am not really sure on how to approach the problem with the given hints, since it would require plotting the graph for different critical values for $a$, which I am not really sure how to find. Thus, I am wondering if the following is correct.
The system $\ddot{x}=(x-a)(x^2-a)$ can be re-written as
$$\begin{cases}
\dot{x}=y\\
\dot{y}=(x-a)(x^2-a)
\end{cases}$$
with fixed points $P_1(a,0),\,P_2(\sqrt{a},0)$ and $P_3(\sqrt{a},0)$. 
The Jacobian is 
$$J(x,y)=\begin{bmatrix}
0 &1\\
3x^2-2ax-a &0
\end{bmatrix}$$
and thus
$$J(a,0)=\begin{bmatrix}
0 &1\\
a^2-a &0
\end{bmatrix};\quad J(\sqrt{a},0)=\begin{bmatrix}
0 &1\\
2a-2a^{3/2} &0
\end{bmatrix}; \quad J(-\sqrt{a},0)=\begin{bmatrix}
0 &1\\
2a+2a^{3/2} &0
\end{bmatrix}.$$
It can be noticed that the $\Tr\left[J(x,y)\right]=0$ and that 
$$\begin{aligned}
&1.\,\Det\left[J(a,0)\right]=a(1-a)\implies \text{Saddle for }a<0 \wedge a>1, \text{Center for }0<a<1.\\
&2.\,\Det\left[J(\sqrt{a},0)\right]=2a(\sqrt{a}-1)\implies \text{Saddle for }0<a<1, \text{Center for }a>1.\\
&3.\,\Det\left[J(-\sqrt{a},0)\right]=-2a(\sqrt{a}+1)\implies \text{Saddle for }a>0.
\end{aligned}$$
If $a=0$ the system reduces to $\ddot{x}=x^3$ where the only fixed point is at $(0,0)$ and thus it is unstable. On the other hand, if $a=1$ then the system reduces to $\ddot{x}=x^3-x^2-x+1$ with fixed points at $(1,0),(-1,0)$, both being unstable. 
Is my work correct?
 A: Almost everything looks right.
For $a > 0$, $-2a < 0$ and $\sqrt{a}+1 > 0$, so $\Det[J(-\sqrt{a},0)] < 0$. This equilibrium should be a saddle then for positive $a$.
Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $x = a, x = \sqrt{a}, x = -\sqrt{a}$ can be graphed on the $ax$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.

This is the bifurcation diagram for your system. The lower black curve is the $x=-\sqrt{a}$ equilibrium, which doesn't have any intersections for $a>0$, so there will be no change in behavior.

Edit:
This isn't specifically asked about in the question but it is an important feature of this equation. For $a>0, a\neq 1$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide  with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $a = \frac{1}{2}$.

