Question

I need help with this problem from my trigonometry prep guideIt asks to solve (a) and (b) Please put these separately so I know which is which

Answer 1

Part 1. We are given the following expression:

$(3x^5-\frac{1}{9}y^3)^4$

This is an expression of the form:

$(a+b)^n$

This expression can be expanded using the binomial theorem as follows:

$(a+b)^n=\sum ^n_{k\mathop=0}\frac{n!}{(n-k)!k!}a^{n-k}b^k$

In this case, we have:

$\begin{gathered} a=3x^5 \\ b=-\frac{1}{2}y^3 \\ n=4 \end{gathered}$

Substituting in the binomial theorem:

$(3x^5-\frac{1}{2}y^3)^4=\sum ^4_{k\mathop{=}0}\frac{4!}{(4-k)!k!}(3x^5)^{4-k}(-\frac{1}{2}y^3)^k$

Part 2. To determine the terms of the binomial we will expand the sum. To do that we will use the values of "k" from 0 to 4:

$\sum ^4_{k\mathop{=}0}\frac{4!}{(4-k)!k!}a^{4-k}b^k=\frac{4!}{(4-0)!0!}a^{4-0}b^0+\frac{4!}{(4-1)!1!}a^{4-1}b^1+\frac{4!}{(4-2)!2!}a^{4-2}b^2+\frac{4!}{(4-3)!3!}a^{4-3}b^3+\frac{4!}{(4-4)!4!}a^{4-4}b^4$

Now, we solve the coefficients and the exponents:

$\sum ^4_{k\mathop{=}0}\frac{4!}{(4-k)!k!}a^{4-k}b^k=a^4+4a^3b^{}+6a^2b^2+4a^{}b^3+b^4$

Now, we substitute the values of "a" and "b":

$\sum ^4_{k\mathop{=}0}\frac{4!}{(4-k)!k!}(3x^5)^{4-k}(-\frac{1}{2}y^3)^k=(3x^5)^4+4(3x^5)^3(-\frac{1}{2}y^3)+6(3x^5)^2(-\frac{1}{2}y^3)^2+4(3x^5)(-\frac{1}{2}y^3)^3+(\frac{1}{2}y^3)^4$

Now, we simplify each term:

$\sum ^4_{k\mathop{=}0}\frac{4!}{(4-k)!k!}(3x^5)^{4-k}(-\frac{1}{2}y^3)^k=81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}$

And thus we get the simplified terms.