Question

I need help with this practice problem It asks to answer (a) & (b) Please put (a) & (b) separately so I know which one is which

Answer 1

The Binomial Theorem

It describes the algebraic expansion of powers of a binomial.

The general formula for the expansion of the binomial is:

$(a+b)^n=\sum ^{k=n}_{k=0}\binom{n}{k}a^kb^{n-k}$

Given the expression:

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

This corresponds to the formula written above for the values:

$a=3x^5,b=-\frac{1}{9}y^3,n=4$

(a) The expression can be written in summation form as:

$(3x^5-\frac{1}{9}y^3)^4=\sum ^{k=4}_{k=0}\binom{4}{k}(3x^5)^k\mleft(-\frac{1}{9}y^3\mright)^{4-k}$

(b) It's required to write the full expansion. We expand the summation as follows (the left part will be omitted for limitations of space):

$\begin{gathered} \binom{4}{0}(3x^5)^0\mleft(-\frac{1}{9}y^3\mright)^4+\binom{4}{1}(3x^5)^1(-\frac{1}{9}y^3)^3+\binom{4}{2}(3x^5)^2(-\frac{1}{9}y^3)^2+ \\ +\binom{4}{3}(3x^5)^3(-\frac{1}{9}y^3)^1+\binom{4}{4}(3x^5)^4(-\frac{1}{9}y^3)^0 \end{gathered}$

Operating:

$\begin{gathered} 1\cdot(1)^{}(\frac{1}{6561}y^{12})^{}+4(3x^5)(-\frac{1}{729}y^9)+6(9x^{10})(\frac{1}{81}y^6)+ \\ +4(27x^{15})(-\frac{1}{9}y^3)+1\cdot(81x^{20})(1) \end{gathered}$

Simplifying:

$\frac{y^{12}}{6561}^{}-\frac{4x^5y^9}{243}+\frac{2x^{10}y^6}{3}-12x^{15}y^3+81x^{20}$