Mohammed TaiyabSiddiqui
Answers!
Samira Singhania
Jan. 27, 2018, 10:35 a.m.
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
Answer the question...
Samira Singhania
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
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