Question 1209937
.
Let f be a function such that
f(xy) + x = xf(y) + f(x) + xy^2
for all real numbers x and y.  If f(-1) = 3, then compute f(100).
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        Below  I  will show,  in couple of lines,  that the posed  " problem " 

        is non-sensical and mathematically  IMPOSSIBLE  self-CONTRADICTORY  gibberish.



Indeed,  take   x =1   in this given identity 


        f(xy) + x = xf(y) + f(x) + xy^2


You will get


        f(y) + 1 = f(y) + f(1) + y^2.


Cancel  f(y)  in both sides and get


        1 = f(1) + y^2,


or


        1 - f(1) = y^2.



It says that the function   y --> y^2   has a constant value of   1 - f(1), 


                which is    I M P O S S I B L E.



Therefore,  I call this  " problem "  as  GIBBERISH:  such  a function,
as described in the post,  does not exist and can not exist.



An immediate consequence from this my post that the  " solution "  by  @CPhill in his post is  GIBBERISH,  too.