Rewriting $\dot{x}=-\bigg[\bigg(x-\frac{1}{2}+\frac{1}{2}\bigg)\bigg(x-\frac{1}{2}-\frac{1}{2}\bigg)+\beta\bigg]2\bigg(x-\frac{1}{2}\bigg)$ I am given the following equation
$$\dot{x}=-\bigg[\bigg(x-\frac{1}{2}+\frac{1}{2}\bigg)\bigg(x-\frac{1}{2}-\frac{1}{2}\bigg)+\beta\bigg]2\bigg(x-\frac{1}{2}\bigg)$$
Which can be  rewritten as
$$=-2\bigg(x-\frac{1}{2}\bigg)^3-2\bigg(\beta-\frac{1}{4}\bigg)\bigg(x-\frac{1}{2}\bigg)$$
I just do not see how. Are there any god tips / hints as to how to reach the final expresion? Any help is highly appreciated :-)
 A: Just use the identity $(a-b)(a+b)=a^{2}-b^{2}$ with $a=x-\frac 1 2, b=\frac 1 2 $ .
A: Write $t=x-1/2$
Then you have:
 $$-[(t+1/2)(t-1/2)+\beta]2t =-[t^2-1/4+\beta]2t = -2t^3-2t(\beta -1/4)$$
A: We have $$\dot{x} = -\left[\left(\color{red}{x-{1\over 2}}\color{blue}{+{1\over 2}}\right)\left(\color{red}{x-{1\over 2}}\color{blue}{-{1\over 2}}\right)+\beta\right]2\left(\color{red}{x-{1\over 2}}\right)$$
remember that $(a+b)(a-b)=a^2-b^2$ and you get $$\dot{x}=-\left[\left(\color{red}{x-{1\over 2}}\right)^2-\color{blue}{1\over 4}+\beta\right]2\left(\color{red}{x-{1\over 2}}\right)$$ now you can easily use distributive property of product and associative property of summation to get $$\dot{x}=-\left[\left(\color{red}{x-{1\over 2}}\right)^2-\left(\color{blue}{1\over 4}-\beta\right)\right]2\left(\color{red}{x-{1\over 2}}\right) = -2\left(\color{red}{x-{1\over 2}}\right)^3-2\left(-\color{blue}{1\over 4}+\beta\right)\left(\color{red}{x-{1\over 2}}\right)$$
A: Hint:let $y=x-\dfrac{1}{2}$ therefore$$y'=-(y^2-\dfrac{1}{4}+\beta)2y$$or $$\dfrac{y'}{(y^2-\dfrac{1}{4}+\beta)2y}=-1$$
