Question

I need help solving this practice problem If you can , answer (a) and (b) separately so I can tell which is which

Answer 1

Step 1:

Write the expression

`(3x^5\text{ - }(1)/(9)y^3)^4`

Step 2:

a)

`\begin{gathered} (3x^5\text{ - }(1)/(9)y^3)^4 \\ =^4C_0(3x^5)^4(-(1)/(9)y^3)^0+^4C_1(3x^5)^3(-(1)/(9)y^3)^1+^4C_2(3x^5)^2(-(1)/(9)y^3)^2+ \\ +^4C_1(3x^5)^1(-(1)/(9)y^3)^3+^4C_0(3x^5_{})^0(-(1)/(9)y^3)^4 \end{gathered}`

Step 3:

b) simplified terms of the expression

`\begin{gathered} Note\colon \\ ^4C_0\text{ = 1} \\ ^4C_1\text{ = 4} \\ ^4C_2\text{ = 6} \\ ^4C_3\text{ = 4} \\ ^4C_4\text{ = 1} \end{gathered}`

Next, substitute in the expression

`\begin{gathered} =\text{ 1}*81x^(20)*1\text{ - 4}*27x^(15)\text{ }*\text{ }(y^3)/(9)\text{ + 6 }*9x^(10)*(y^6)/(81)\text{ - 4}*3x^5\text{ }*\text{ }(y^9)/(729) \\ +\text{ 1 }*\text{ 1 }*(y^(12))/(6561)\text{ } \end{gathered}`
`=81x^(20)-12x^(15)y^3\text{ + }(2)/(3)x^(10)y^6\text{ - }(4)/(243)x^5y^9\text{ + }(1)/(6561)y^(12)`