Chapter 9.2, Problem 26E

Solution

To identify: The expansion of the binomial ${\left(a-b^{-3}\right)}^7$ by using the binomial theorem.

The expanded form of the binomial is $a^7-\frac{7a^6}{b^3}+\frac{21a^5}{b^6}-\frac{35a^4}{b^9}+\frac{35a^3}{b^{12}}-\frac{21a^2}{b^{15}}+\frac{7a}{b^{18}}-\frac{1}{b^{21}}$

Given information:

The given expression is ${\left(a-b^{-3}\right)}^7$ .

Explanation:

Consider the given expression

The binomial Theorem state that for any positive integer

  $n,$ ${\left(a+b\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)a^n+\left(\begin{array}{l}n\\ 1\end{array}\right)a^{n-1}b+\mathrm{......}+\left(\begin{array}{l}n\\ r\end{array}\right)a^{n-r}{b}^r+\mathrm{.....}+\left(\begin{array}{l}n\\ n\end{array}\right)b^n$

Where $a=a,n=7\text{and}b=-b^{-3}$

  ${\left(a+-b^{-3}\right)}^7=\left(\begin{array}{l}7\hfill \\ 0\hfill \end{array}\right){\left(a\right)}^7{\left(-b^{-3}\right)}^0+\left(\begin{array}{l}7\hfill \\ 1\hfill \end{array}\right){\left(a\right)}^6{\left(-b^{-3}\right)}^1+\left(\begin{array}{l}7\hfill \\ 2\hfill \end{array}\right){\left(a\right)}^5{\left(-b^{-3}\right)}^2+\left(\begin{array}{l}7\hfill \\ 3\hfill \end{array}\right){\left(a\right)}^4{\left(-b^{-3}\right)}^3+\left(\begin{array}{l}7\hfill \\ 4\hfill \end{array}\right){\left(a\right)}^3{\left(-b^{-3}\right)}^4$

  $+\left(\begin{array}{l}7\hfill \\ 5\hfill \end{array}\right){\left(a\right)}^2{\left(-b^{-3}\right)}^5+\left(\begin{array}{l}7\hfill \\ 5\hfill \end{array}\right){\left(a\right)}^1{\left(-b^{-3}\right)}^6+\left(\begin{array}{l}7\hfill \\ 5\hfill \end{array}\right){\left(a\right)}^0{\left(-b^{-3}\right)}^7$

  ${\left(a+-b^{-3}\right)}^7=a^7-7a^6{b}^{-3}+21a^5{b}^{-6}-35a^4{b}^{-9}+35a^3{b}^{-12}-21a^2{b}^{-15}+7a{b}^{-18}-b^{-21}$

  ${\left(a+-b^{-3}\right)}^7=a^7-\frac{7a^6}{b^3}+\frac{21a^5}{b^6}-\frac{35a^4}{b^9}+\frac{35a^3}{b^{12}}-\frac{21a^2}{b^{15}}+\frac{7a}{b^{18}}-\frac{1}{b^{21}}$

So, the expression by using theorem

  $a^7-\frac{7a^6}{b^3}+\frac{21a^5}{b^6}-\frac{35a^4}{b^9}+\frac{35a^3}{b^{12}}-\frac{21a^2}{b^{15}}+\frac{7a}{b^{18}}-\frac{1}{b^{21}}$