Remove anything before certain combination

I want to leave the last part like "word+000" combination, remove anything before:

Like:

“34_ert345wers000” would become "wers000"
"fdsf_grge000" would become "grge000"
"#^%^$%jnrjr000" would become "jnrjr000"

I tried replace method,
Text to be replaced: (.+)(\w*000)
Replace with: $2

But it didn't work properly, the regular expressions will always search from left to right, so it cannot decide (\w*000) first and then decide the left.

You have two problems...

1) The metacharacter \w will match numerical digits and underscore as well as letters
ie \w = [a-zA-Z0-9_]
so you need to explicitly define the characters you wish to match as [a-z]
(or [a-zA-Z] if your regex is case-sensitive).

2) The "+" & "*" modifiers are "greedy" by default so ".+" will match everything - including word characters - up to "000" in your regex. The solution is to append "?" to the modifier, which is an instruction for the modifier to be "lazy".

So...
Replace: .*?([a-z]*000$)
With: \1

Reply to #2:

That's worked out so well, Thank you very much!

Meanwhile I learnt a lot, thank you!