Question

I need help with this practice problem. It’s from my trigonometry prep guide.It asks to answer (a) and (b)**Please put these separately so I know which is which**

Answer 1

a) The binomial theorem states the following:

`(x+y)^n=\sum ^n_(k\mathop=0)(n!)/(k!(n-k)!)x^(n-k)y^k`

Therefore, we have:

`(3x^5-(1)/(9)y^3)^4=\sum ^4_(k=0)(4!)/(k!(4-k)!)3^(4-k)x^(5(4-k))(-(1)/(9))^ky^(3k)`

b) The terms are:

`\begin{gathered} k=0\rightarrow(4!)/(0!(4-0)!)3^(4-0)x^(5(4-0))(-(1)/(9))^0y^(3\cdot0)=81x^(20) \\ k=1\rightarrow(4!)/(1!(4-1)!)3^(4-1)x^(5(4-1))(-(1)/(9))^1y^(3\cdot1)=-12x^(15)y^3 \\ k=2\rightarrow(4!)/(2!(4-2)!)3^(4-2)x^(5(4-2))(-(1)/(9))^2y^(3\cdot2)=(2)/(3)x^(10)y^6 \\ k=3\rightarrow(4!)/(3!(4-3)!)3^(4-3)x^(5(4-3))(-(1)/(9))^3y^(3\cdot3)=-(4)/(81)x^5y^9 \\ k=4\rightarrow(4!)/(4!(4-4)!)3^(4-4)x^(5(4-4))(-(1)/(9))^4y^(3\cdot4)=(1)/(6561)y^(12) \end{gathered}`