a) The binomial theorem states the following:
[tex](x+y)^n=\sum ^n_{k\mathop=0}\frac{n!}{k!(n-k)!}x^{n-k}y^k[/tex]
Therefore, we have:
[tex](3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k=0}\frac{4!}{k!(4-k)!}3^{4-k}x^{5(4-k)}(-\frac{1}{9})^ky^{3k}[/tex]
b) The terms are:
[tex]\begin{gathered} k=0\rightarrow\frac{4!}{0!(4-0)!}3^{4-0}x^{5(4-0)}(-\frac{1}{9})^0y^{3\cdot0}=81x^{20} \\ k=1\rightarrow\frac{4!}{1!(4-1)!}3^{4-1}x^{5(4-1)}(-\frac{1}{9})^1y^{3\cdot1}=-12x^{15}y^3 \\ k=2\rightarrow\frac{4!}{2!(4-2)!}3^{4-2}x^{5(4-2)}(-\frac{1}{9})^2y^{3\cdot2}=\frac{2}{3}x^{10}y^6 \\ k=3\rightarrow\frac{4!}{3!(4-3)!}3^{4-3}x^{5(4-3)}(-\frac{1}{9})^3y^{3\cdot3}=-\frac{4}{81}x^5y^9 \\ k=4\rightarrow\frac{4!}{4!(4-4)!}3^{4-4}x^{5(4-4)}(-\frac{1}{9})^4y^{3\cdot4}=\frac{1}{6561}y^{12} \end{gathered}[/tex]
