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

For the expression

`\begin{gathered} (3x^5-(1)/(9)y^3)^4 \\ =(3x^5+(-(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=-(y^3)/(9),n=4 \end{gathered}`

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

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

Hence the answer is

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

(b) So, expanding we have

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

Continuing the expansion we have

Then the final answer becomes

`81x^(20)-12x^(15)y^3+(2x^(10)y^6)/(3)-(4x^5y^9)/(243)+(y^(12))/(6561)`