Pitchfork bifurcation Pitchfork breaking symmetry
I have this equation and I want to find when the breaking symmetry occurs:
$$
-x^3 + x^2(b - 1) + x(b - a)
$$
The roots are $x=0$ and $-x^2 + x(b - 1) + (b - a)=0$
The other solutions are
$$
\frac{b}{2} - \frac{1}{2} \pm 
\frac{\sqrt{b^2 - 4a + 2b + 1}}{2} 
$$
What I observe is, we have always the solution $x=0$, so if we try to find the breaking symmetry we need the other two solution be equal to zero.
The first condition is when $b=1$ and then the roots become
$\sqrt{1-a}$ and $-\sqrt{1-a}$.
here the picture becomes

Are my steps correct?
Edit:
I would like to thank @Sammy Black.
we want to find where the symmetry breaking occurs.
If we set (x)-->(-x) the equation is not symmetric.
see $f=−^3+^2(−1)+(−).$
Then $f(-x) = x^3+x^2(b-1)-x(b-a)$. The problem is with the quadratic term. So, can we say that the broken symmetry occurs when $b\neq 1$? If b=1, then we lose the quadratic term and we then have the normal form for pitchfork bifurcation.
Thank you.
 A: The two solutions to a quadratic equation
$$
Ax^2 + Bx + C = 0
$$
(with $A \neq 0$) become one solution where the discriminant
$$
\Delta = B^2 - 4AC
$$
is zero. In your equation, once you factor out $x$ (which you correctly associate with the solution $x=0$), the coefficients are
$$
\left\{ 
\begin{aligned} 
A &= -1 \\ 
B &= b - 1 \\ 
C &= b - a
\end{aligned} 
\right.
$$
Now,
$$
\Delta = (b - 1)^2 - 4(-1)(b - a) 
= (b^2 - 2b + 1) + (4b - 4a)
= (b + 1)^2 - 4a
$$
so the bifurcation occurs where the parameters $a$ and $b$ conspire to make $\Delta = 0$, i.e. where
$$
a = \biggl( \frac{b+1}{2} \biggr)^2.
$$
In your bifurcation diagram, you've assumed that $b=1$, so you find the pitchfork bifurcation at $a=1$, but there are actually two parameters, so the bifurcation diagram consists of surfaces graphed over the $(a, b)$-plane. This is more difficult to picture.
For particular values of $b$ in the range $\{-15, \dots, 15\}$, the $x$-equilibrium values as a function of the parameter $a$ look like this:

On the other hand, for particular values of $a$ in the range $\{-15, \dots, 15\}$, the $x$-equilibrium values as a function of the parameter $b$ look like this:

