Question

I need help with this It’s from my trigonometry prep book It asks to answer (a) and (b) Put these separately ^ so I know which is which

Answer 1

SOLUTION

For the expression

$\begin{gathered} (3x^5-\frac{1}{9}y^3)^4 \\ =(3x^5+(-\frac{y^3}{9}))^4 \end{gathered}$

(a) The summation notation that he will use to expand the expression would be

$\begin{gathered} (a+b)^n=\sum ^n_{r\mathop=0}(^n_r)a^{n-r}b^r=\sum ^n_{r\mathop{=}0}(^n_r)a^rb^{n-r} \\ \text{where } \\ a=3x^5,b=-\frac{y^3}{9},n=4 \end{gathered}$

substituting the values of a, b and n, we have

$(3x^5+(-\frac{y^3}{9}))^4=\sum ^4_{r\mathop{=}0}(^4_r)(3x^5)^r(-\frac{y^3}{9})^{4-r}$

Hence the answer is

$(3x^5+(-\frac{y^3}{9}))^4=\sum ^4_{r\mathop{=}0}(^4_r)(3x^5)^r(-\frac{y^3}{9})^{4-r}$

(b) So, expanding we have

$\begin{gathered} \frac{4!}{0!\left(4-0\right)!}\mleft(3x^5\mright)^4\mleft(-\frac{y^3}{9}\mright)^0+\frac{4!}{1!\left(4-1\right)!}\mleft(3x^5\mright)^3\mleft(-\frac{y^3}{9}\mright)^1+ \\ \frac{4!}{0!\left(4-0\right)!}\mleft(3x^5\mright)^4\mleft(-\frac{y^3}{9}\mright)^0+\frac{4!}{1!\left(4-1\right)!}\mleft(3x^5\mright)^3\mleft(-\frac{y^3}{9}\mright)^1+ \\ \frac{4!}{3!\left(4-3\right)!}\mleft(3x^5\mright)^1\mleft(-\frac{y^3}{9}\mright)^3+\frac{4!}{4!\left(4-4\right)!}\mleft(3x^5\mright)^0\mleft(-\frac{y^3}{9}\mright)^4 \end{gathered}$

Continuing the expansion we have

Then the final answer becomes

$81x^{20}-12x^{15}y^3+\frac{2x^{10}y^6}{3}-\frac{4x^5y^9}{243}+\frac{y^{12}}{6561}$