## Precalculus (6th Edition) Blitzer

$56x^9y^{10}$
Calculate the sixth term for for $(x^3+y^2)^{8}$ by using General formula such as:$(m+n)^r=\displaystyle \binom{r}{k}m^{r-k}n^k$ and $\displaystyle \binom{r}{k}=\dfrac{r!}{k!(r-k)!}$ This implies,$(x^3+y^2)^{8}=\displaystyle \binom{8}{3}(x^3)^{8-5}(y^2)^5$ or,$=\dfrac{8!}{5!(8-5)!}x^{9}y^{10}$ or, $=[\dfrac{ 8 \times 7 \times 6 \times 5!}{5 !(3 \times 2 \times 1)}]x^{9}y^{10}$ or, $=56x^9y^{10}$