Chapter 7.4, Problem 44PPE

To determine

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

Answer to Problem 44PPE

The required answer is: $-\frac{375842.29y^{12}}{x^{\frac{1}{2}}}$

Explanation of Solution

Given information:

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

Calculation:

  ${\left(-\frac{2x^{\frac{1}{6}}}{3y^4}\right)}^{-3}=\frac{1}{{\left(-\frac{2x^{\frac{1}{6}}}{3y^4}\right)}^3}$

(Rewrite using positive exponents)

  $=\frac{1}{\left(-\frac{{(2x^{\frac{1}{6}})}^3}{{(3y^4)}^3}\right)}$

(Raise the quotient to third power)

  $=\frac{1}{\left(-\frac{{(2x)}^{\frac{1}{2}}}{{(3y)}^{12}}\right)}$

(Multiply the exponents in the quotient)

  $=1\cdot \left(-\frac{{(3y)}^{12}}{{(2x)}^{\frac{1}{2}}}\right)$

(Multiply by the reciprocal of $\left(-\frac{{(2x)}^{\frac{1}{2}}}{{(3y)}^{12}}\right)$ which is $\left(-\frac{{(3y)}^{12}}{{(2x)}^{\frac{1}{2}}}\right)$ )

  $\begin{array}{l}=-\frac{3^{12}{y}^{12}}{2^{\frac{1}{2}}{x}^{\frac{1}{2}}}\\ \\ =-\frac{531441y^{12}}{\sqrt[2]{2}{x}^{\frac{1}{2}}}\end{array}$

(Replace the value of $2^{\frac{1}{2}}=\sqrt[2]{2}$ )

  $\begin{array}{l}=-\frac{531441y^{12}}{1.414x^{\frac{1}{2}}}\\ \\ =-\frac{375842.29y^{12}}{x^{\frac{1}{2}}}\end{array}$

(Simplify)