
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Given the inequality $ax^{2}+bx+c \gt 0$ with the solution set $\{x\left|\right.1 \lt x \lt 2\}$, find the solution set of the inequality $cx^{2}+bx+a \gt 0$ in terms of $x$. When studying the above problem, Xiaoming and Xiaoning respectively came up with the following Solution 1 and Solution 2:

**Solution 1:** From the given information, the roots of the equation $ax^{2}+bx+c=0$ are $1$ and $2$, and $a \lt 0$. By Vieta's formulas, we have $\left\{\begin{array}{c}1+2=-\frac{b}{a},\\ 1×2=\frac{c}{a},\end{array}\right.\left\{\begin{array}{c}b=-3a,\\ c=2a,\end{array}\right.$. Therefore, the inequality $cx^{2}+bx+a \gt 0$ can be transformed into $2ax^{2}-3ax+a \gt 0$, which simplifies to $\left(x-1\right)\left(2x-1\right) \lt 0$. Solving this inequality gives $\frac{1}{2}＜x＜1$, so the solution set of the inequality $cx^{2}+bx+a \gt 0$ is $\{x|\frac{1}{2}＜x＜1\}$.

**Solution 2:** From $ax^{2}+bx+c \gt 0$, we get $c{(\frac{1}{x})}^{2}+b\frac{1}{x}+a＞0$. Let $y=\frac{1}{x}$, then $\frac{1}{2}＜y＜1$. Therefore, the solution set of the inequality $cx^{2}+bx+a \gt 0$ is $\{x|\frac{1}{2}＜x＜1\}$. 

Based on the above solutions, answer the following questions:

$(1)$ If the solution set of the inequality $\frac{k}{x+a}+\frac{x+c}{x+b}＜0$ is $\{x\left|\right.-2 \lt x \lt -1$ or $2 \lt x \lt 3\}$, write down the solution set of the inequality $\frac{kx}{ax+1}+\frac{cx+1}{bx+1}＜0$ directly.

$(2)$ If real numbers $m$ and $n$ satisfy the equations $\left(m+1\right)^{2}+\left(4m+1\right)^{2}=1$ and $\left(n+1\right)^{2}+\left(n+4\right)^{2}=n^{2}$, and $mn\neq 1$, find the value of $n^{3}+m^{-3}$.

Present the answer in LaTex format: \boxed{Your answer}
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