The correct option is B 476
We have,
[1+x2(1−x)]8=(1+x2−x3)8=∑k1+k2+k3=88!k1!k2!k3!(1)k1(x2)k2(−x3)k3=∑k1+k2+k3=88!k1!k2!k3!(−1)k3x2k2+3k3
Coefficient of x10, so
2k2+3k3=10k1+k2+k3=8
So,
(1) k1=4,k2=2,k3=2(2) k1=3,k2=5,k3=0
Therefore, the required coefficient
=8!4!2!2!+8!3!5!0!=8⋅7⋅6⋅54+8⋅7⋅66=420+56=476