Answer with Step-by-step explanation:
We are given that
[tex](3x^5-\frac{1}{9}y^3)^4[/tex]
a.We know that binomial theorem in summation form
[tex](a+b)^n=\sum_{r=0}^{r=n}\binom{n}{r}a^{n-r}\cdot b^r[/tex]
Using the formula and substitute [tex]a=3x^5,b=-\frac{1}{9}y^3[/tex] and n=4
[tex](3x^5-\frac{1}{9}y^3)^4=\sum_{r=0}^{r=4}\binom{4}{r}(3x^5)^{4-r}\cdot (-\frac{1}{9}y^3)^r[/tex]
b.[tex](3x^5-\frac{1}{9}y^3)^4=\sum_{r=0}^{r=4}(3x^5)^{4-r}(-\frac{1}{9}y^3)^r=\binom{4}{0}(3x^5)^4+\binom{4}{1}(3x^5)^3(-\frac{1}{9}y^3)+\binom{4}{2}(3x^5)^2(-\frac{1}{9}y^3)^2+\binom{4}{3}(3x^5)(-\frac{1}{9}y^3)^3+\binom{4}{4}(-\frac{1}{9}y^3)^4[/tex]
Combination formula:[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
[tex]n!=n(n-1)(n-2)...3\cdot 2\cdot 1[/tex]
[tex](3x^5-\frac{1}{9}y^3)^4=\frac{4!}{0!4!}(81x^{20})+\frac{4!}{3!}(27x^{15})(-\frac{1}{9}y^3)+\frac{4!}{2!2!}(9x^{10})(\frac{1}{81}y^6)+\frac{4!}{3!}(3x^5)(-\frac{1}{729}y^9)+\frac{4!}{4!}(\frac{1}{6561}y^{12})[/tex]
[tex](3x^5-\frac{1}{9}y^3)^4=81x^{20}-\frac{4\times 3!}{3!}(3x^{15}y^3+\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{1}{9}x^{10}y^6-\frac{4\times 3!}{3!}(\frac{x^5y^9}{243}+\frac{6561}y^{12}[/tex]
[tex](3x^5-\frac{1}{9}y^3)^4=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]
