How do you solve and write the following interval rotation: $x\left( x-1 \right)\left( x+3 \right)>0?$

Hint: In interval notation we just write the beginning and ending numbers of the interval. When we write interval rotation $\left[ {} \right]$ this bracket is used when we want to include the end value and use $\left( {} \right)$ a round bracket when we don’t want to include the end value. We have to solve the inequality.
There are different methods of solving inequality such as by algebraic method. By the number line method and graphing method.

Complete step-by-step answer:
We have to solve the inequality $x\left( x-1 \right)\left( x+3 \right)>0$ from this we know that the product $x\left( x-1 \right)\left( x+3 \right)$ is positive.
It is also apparent that signs of binomial $\left( x+ \right)$ and $\left( x-1 \right)$ will change ground value $-3.9$ and respectively. We divide the real number line using around these values i.e. below $-3,$ between $-3$and $0$.
Between $0$ and $1$ and above $1$ and see how the sign changes, as we move across the real number line.


It is observed that $x\left( x-1 \right)\left( x+3 \right)>0$ when either -3 < x < 0 or x > 1 this is the solution for the inequality.
This can be written in interval rotation as $\left( -3,0 \right)$ or $\left( 1,\infty \right)$
Here small brackets $\left( {} \right)$ indicates the end points are not included through the solution including values in the interval.
This is used when we have only inequality.
If we have equality included, say for $x\left( x-1 \right)\left( x+3 \right)$ we use square brackets which means end points are included.

So, the solution in interval rotation is $\left( -3,0 \right)$ as $\left( 1,\infty \right)$

Additional Information:
There are three methods for solving inequalities.
Explanation:
First method: By algebraic method
Example: solve: 2x - 7 < x - 5
2x - x < 7 - 5
$x < 2$
Second method: By number line method:
Example solve $f(x)={{x}^{2}}+2x-3<0$
First, solve $f(x)=0$ There are $2$ real roots ${{x}_{1}}=1\ And {{x}_{2}}=-3$
Replace $x=0$ into $f(x)$ we find $E(0)=-3<0$
Therefore, the origin $'0'$ is located inside the solution set
Answer by interval: $\left( -3,1 \right)$
By graphing method: Example: Solve $f(x)={{x}^{2}}+2x-3<0$
The graph of $f(x)$ is an upward parabola $\left( a>0 \right),$ that intersects the $x$-axis at ${{x}_{1}}=1$ and ${{x}_{2}}=-3$. Inside the interval $\left( -3,1 \right)$ the parabola stay below the $ay>>f(x)<0$
Therefore, the solution set is the open interval $\left( -3,1 \right)$

Note:
While writing interval rotation carefully give the bracket if you are considering or including the end value give $\left[ {} \right]$ square bracket and if you are not including end point give $\left( {} \right)$ square bracket.