Good evening forum I am having a problem with the following problem any help would be appreciated:
thank you , AC
use the binomial expansion formula for $\displaystyle (x + y)^8$
we have $\displaystyle (x + y)^8 = \sum_{k = 0}^8 {8 \choose k}x^{8 - k}y^k$
if we choose $\displaystyle x = a$ and $\displaystyle y = -b^3$ we realize that we want the term where we have $\displaystyle y^3$...that's where k = 3
this is the term $\displaystyle {8 \choose 3}x^5y^3$
note that $\displaystyle {n \choose k} = _nC_k = \frac {n!}{k!(n - k)!}$