Answers!
For n odd integer the following identity holds
xn+yn=(x+y)(xn−1−xn−2y+⋯−xyn−2+yn−1)
or
xn+ynx+y=(xn−1−xn−2y+⋯−xyn−2+yn−1)
Hence for our case we have that
x5+y5x+y=x4−x3y+x2y2−xy3+y4
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Answer the question...
For n odd integer the following identity holds
xn+yn=(x+y)(xn−1−xn−2y+⋯−xyn−2+yn−1)
or
xn+ynx+y=(xn−1−xn−2y+⋯−xyn−2+yn−1)
Hence for our case we have that
x5+y5x+y=x4−x3y+x2y2−xy3+y4
Comment