We have,
(1−x)101(x2+x+1)100
=(1−x)100(x2+x+1)100(1−x)
=[(1−x)(x2+x+1)]100⋅(1−x)
=(1−x3)100⋅(1−x)
=Notermofx256⏟(1−x3)100−wefindcofficientofx255⏟x(1−x3)100
Now,
(1−x3)100=C0100−C1100x3+C2100x6−...−C85100(−x255)
⇒x(1−x3)100=C0100x−C1100x4+C2100x7−...+C85100x256
Required coefficient
=C85100=C15100