We need to expand the following binomial:
[tex]\mleft(3x^5-\frac{1}{9}y^3\mright)^4[/tex]
using binomial theorem. This theorem states that:
[tex]\mleft(a+b\mright)^n=\sum ^n_{i=0}\binom{n}{i}a^{\mleft(n-i\mright)}b^i[/tex]
where
[tex]\binom{n}{i}=\frac{n!}{i!(n-i)!}[/tex]
In this problem, we have:
[tex]\begin{gathered} a=3x^5 \\ \\ b=-\frac{1}{9}y^{3} \\ \\ n=4 \end{gathered}[/tex]
Part (a)
To express the expansion, we use the following sum in summation notation:
[tex]\mleft(3x^5-\frac{1}{9}y^3\mright)^4=\sum ^4_{i=0}\binom{4}{i}(3x^5)^{(4-i)}\mleft(-\frac{1}{9}y^3\mright)^{^i}[/tex]
This can also be written as:
[tex]\mleft(3x^5-\frac{1}{9}y^3\mright)^4=\sum ^4_{i=0}\frac{4!}{i!(4-i)!}(3x^5)^{(4-i)}\mleft(-\frac{1}{9}y^3\mright)^{^i}[/tex]
Part (b)
Simplifying the terms, we obtain:
[tex]\begin{gathered} \frac{4!}{0!4!}\cdot81x^{20}+\frac{4!}{1!3!}\cdot27x^{15}\cdot(-\frac{1}{9})y^3+\frac{4!}{2!2!}\cdot9x^{10}\cdot\frac{1}{81}y^6 \\ \\ +\frac{4!}{3!1!}\cdot3x^5\cdot(-\frac{1}{729})y^9+\frac{4!}{4!0!}\cdot\frac{1}{6561}y^{12} \\ \\ =81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12} \\ \end{gathered}[/tex]
Therefore, the simplified terms of the expansion are:
[tex]81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}[/tex]
