Answer
$56x^9y^{10}$
Work Step by Step
Our aim is to find the sixith term for $(x^3+y^2)^{8}$.
General formula:$(p+q)^n=\displaystyle \binom{n}{k}p^{n-k}q^k$
and $\displaystyle \binom{n}{k}=\dfrac{n!}{k!(n-k)!}$
Now, $(x^3+y^2)^{8}=\displaystyle \binom{8}{3}(x^3)^{8-5}(y^2)^5$
This implies,
$=\dfrac{8!}{5!(8-5)!}(x)^{9}(y)^{10}$
$=\dfrac{ 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{5 \times 4 \times 3 \times 2 \times 1(3!)}(x)^{9}y^{10}$
Thus,
$=56x^9y^{10}$
