View Single Post
Old 2021-07-29, 19:24   #14
RomanM
 
Jun 2021

23×5 Posts
Default

) we start from (b-y)^2? For cube, (b-y)^3 must be expanded (b^3-3*b^2*y+3*b*y^2-y^3) and solved in whole by the same manner, without any simplification from the start. And I'm stuck with imaginary parts and 3-roots)) At the same time, cube and highger have solution, just like square!
RomanM is offline   Reply With Quote