@GabrielRomon The asker regularly posts PSQs. The question you are referring to is one such PSQ, with no effort shown (do you see any?). Another one is here. If you think it is not a PSQ then nothing on Math SE is a PSQ.
There is not feature on the site to write to a user directly. One can contact users using the chat feature, via inviting them to a chat (if they have a chatprofile). For the specific case you could ask in chat.stackexchange.com/rooms/2165/crude where some of the involved users are around frequently. (In fact chances are the deletion originated from a message there.) — quid ♦12 hours ago
As quid pointed out, this deleted question is a one of the listed question in this comment. — John Ma9 hours ago
@user21820 these are low quality questions indeed, but I don't think they deserve deletion given the extra efforts of Jack d'Aurizio and Blue to produce Latex geometric figures
@GabrielRomon Just because people spend effort on poor questions does not make the questions any less poor. They can and should post their answers on good questions. If a poor question gets deleted and someone thinks their answer is applicable to a much wider audience than the original (lousy) one, they can just post their own (often generalized) question and answer independently and let the original be deleted.
I'll also note that your method of undeleting and locking to preserve a good answer seems to be sub-optimal. As Martin and I have discussed extensively before, there is another option available (which Martin has in fact used a number of times): Write a new question (a different one) and move the good answer there, leaving the old lousy question closed. Jam can do this, or you (as a moderator) can do it if you love Jam's answer.
How can you simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
I know that $\cos^2x+\sin^2x=1$. I thought about trying to use $(\cos^2x+\sin^2x)^3=1$, but that did not lead me anywhere.
I found some other questions about the same expression, but they simplify this to another form: Finding $\si...
How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $(\cos^2x+\sin^2x)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparin...
@GabrielRomon: So as you can see, in both cases this approach is not only fully in line with the goal of preserving question quality on Math SE, but also the moderators are more than happy to help to move quality answers to a good substitute for a bad question. Furthermore, Martin will be able to tell you that this approach is rather well received (even though Martin says that sometimes there will still be inexplicable downvotes on the new good question).
I guess you can ask @MartinSleziak if you wish to know his viewpoint (which differs from mine in certain respects), though we both find this approach acceptable.
@amWhy Incredible. I strongly believe that anyone who has really gone through "a 100 book" as the asker claims, would have found at least 1 good book to be able to ask a better question that that; right now I can't even guess what it's about.
@user21820 <− @amWhy @Did @SimplyBeautifulArt: Just 1 more vote needed.
@amWhy @Did: It is closed and ready for deletion (no doubt due to showing no effort).
@SimplyBeautifulArt It's just false, and anyone who has a rudimentary understanding of even the mere statement of the halting problem and unprovability should know.
@maxematician: Hello! I saw your most recent comment on your question, and have decided to vote to reopen. Please be assured that those of us who may appear harsh with poor questions do actually love to help people who actually want to learn mathematics!
Also, this question, (and the previous three questions by the same asker) math.stackexchange.com/questions/2527801/… needs to be closed. (Take a peek at the asker's other questions, some of which I've voted to close, if you have the time.).