Chapter 7.4, Problem 75PPE

To determine

To simplify: the expression ${\left(\frac{a^{-1}{b}^{3}c}{a^2{b}^4}\right)}^6$

Answer to Problem 75PPE

The required answer is: $\frac{c^6}{a^{18}{b}^6}$

Explanation of Solution

Given information:

The given expression: ${\left(\frac{a^{-1}{b}^{3}c}{a^2{b}^4}\right)}^6$

Calculation:

  ${\left(\frac{a^{-1}{b}^{3}c}{a^2{b}^4}\right)}^6=\frac{{\left(a^{-1}{b}^{3}c\right)}^6}{{\left(a^2{b}^4\right)}^6}$

(Raise the numerator and denominator to the sixth power)

  $=\frac{{(a^{-1})}^6{(b^3)}^6{c}^6}{{(a^2)}^6{(b^4)}^6}$

(Multiply the exponents in the numerator and denominator)

  $\begin{array}{l}=\frac{a^{-6}{b}^{18}{c}^6}{a^{12}{b}^{24}}\\ =a^{-6-12}{b}^{18-24}{c}^6\end{array}$

(Subtract exponents when dividing powers with the same base)

  $\begin{array}{l}=a^{-18}{b}^{-6}{c}^6\\ =\frac{1}{a^{18}}\cdot \frac{1}{b^6}{c}^6\end{array}$

(Rewrite using positive exponents)

  $=\frac{c^6}{a^{18}{b}^6}$

(Simplify)