
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 that the solution set of the inequality $ax^2+bx+c>0$ is $(1,2)$, solve the inequality $cx^2+bx+a>0$.

Here is a solution method:  
**Solution:** Since the solution set of $ax^2+bx+c>0$ is $(1,2)$, we get that the solution set of $a\left( \frac {1}{x}\right)^2+b\left( \frac {1}{x}\right)+c>0$ is $\left( \frac {1}{2},1\right)$, which means the solution set of the inequality $cx^2+bx+a>0$ is $\left( \frac {1}{2},1\right)$.

Based on the above solution method: If the solution set of the inequality $\frac {b}{x+a}+ \frac {x+b}{x+c}<0$ is $(-1,-\frac {1}{3})\cup(\frac {1}{2},1)$, then the solution set of the inequality $\frac {b}{x-a}- \frac {x-b}{x-c}>0$ is \_\_\_\_\_\_.

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