Answer:
28
Step-by-step explanation:
Consider the expansion:
[tex]$(x+y)^2=\sum_{k=0}^{n} \binom{n}{k}y^{n-k}x^k$[/tex]
[tex]$ \binom{n}{k}=\frac{n!}{(n-k)!k!}[/tex]
We have
[tex]$\left(x + y\right)^{8}=\sum_{k=0}^{8} \binom{8}{k} \left(y\right)^{8-k} \left(x\right)^k[/tex]
I'm sorry but I will not write and solve everything, typing LaTeX takes time. But that's the idea:
For [tex]k=0[/tex]
[tex]$\binom{8}{0} \left(y\right)^{8-0} \left(x\right)^{0}=\frac{8!}{(8-0)! 0!}\left(y\right)^{8} \left(x\right)^{0}=y^{8}[/tex]
For [tex]k=2[/tex]
[tex]$\binom{8}{2} \left(y\right)^{8-2} \left(x\right)^{2}=\frac{8!}{(8-2)! 2!}\left(y\right)^{6} \left(x\right)^{2}=28 x^{2} y^{6}[/tex]
[tex](x+y)^8={x}^{8} + 8{x}^{7}y + 28{x}^{6}{y}^{2} + 56{x}^{5}{y}^{3} + 70{x}^{4}{y}^{4} + 56{x}^{3}{y}^{5} + 28{x}^{2}{y}^{6} + 8x{y}^{7} + {y}^{8}[/tex]
