I had a function:
$$\frac{\left(x^3-3x\right)}{\left((x^2+2x+1)^{\left(\frac{1}{\left(2\right)}\right)}\right)}$$
I noted that the it can be re-written as:
$$\frac{\left(x^3-3x\right)}{x+1}$$
Because it is root of a square of a binomial, but I do not understand why the graphs of the first and the second are different for $x < -1$
Why is this simplification wrong?
For the graphs refer to the below link:
On
$$\frac { \left( x^{ 3 }-3x \right) }{ \sqrt { { x }^{ 2 }+2x+1 } } =\frac { x\left( x-\sqrt { 3 } \right) \left( x+\sqrt { 3 } \right) }{ \sqrt { { \left( x+1 \right) }^{ 2 } } } =\frac { x\left( x-\sqrt { 3 } \right) \left( x+\sqrt { 3 } \right) }{ \left| x+1 \right| } =\begin{cases} \frac { x\left( x-\sqrt { 3 } \right) \left( x+\sqrt { 3 } \right) }{ x+1 } ,x+1>0 \\ -\frac { x\left( x-\sqrt { 3 } \right) \left( x+\sqrt { 3 } \right) }{ x+1 } ,x+1<0 \end{cases}$$
Your issue arises simply because $$\sqrt{(x+1)^2} = |x+1|$$ and $x+1 \neq |x+1|$ when $x<-1$.