Question

I need help with this practice problem.It’s from my trig prep guide. It asks to answer (a) & (b).

Answer 1

The general binomial theorem can be expressed as:

`(a+b)^n=\sum ^n_(k\mathop=0)C^n_k\cdot a^(n-k)\cdot b^k`

Now, for this problem we identify:

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

(a)

Then, using the general form:

`(3x^5-(1)/(9)y^3)^4=\sum ^4_{k\mathop{=}0}C^4_k\cdot(3x^5)^(4-k)\cdot(-(1)/(9)y^3)^k`

(b)

The combination operator for this sum:

`\begin{gathered} C^4_0=C^4_4=1 \\ C^4_1=C^4_3=4 \\ C^4_2=6 \end{gathered}`

Then, the simplified terms of the expansion are:

`\begin{gathered} C^4_0\cdot(3x^5)^4\cdot(-(1)/(9)y^3)^0=81x^(20) \\ C^4_1\cdot(3x^5)^3\cdot(-(1)/(9)y^3)^1=4\cdot(27x^(15))\cdot(-(1)/(9)y^3)=-12x^(15)y^3 \\ C^4_2\cdot(3x^5)^2\cdot(-(1)/(9)y^3)^2=6\cdot(9x^(10))\cdot((1)/(81)y^6)=(2)/(3)x^(10)y^6 \\ C^4_3\cdot(3x^5)^1\cdot(-(1)/(9)y^3)^3=4\cdot(3x^5)\cdot(-(1)/(729)y^9)=-(4)/(243)x^5y^9 \\ C^4_4\cdot(3x^5)^0\cdot(-(1)/(9)y^3)^4=(1)/(6561)y^(12) \end{gathered}`