Answer
$56x^9y^{10}$
Work Step by Step
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, $=56x^9y^{10}$
