Chapter 6.2, Problem 63WE

To determine

To Simplify:

  $\sqrt[3]{6y^{-4}}\cdot \sqrt[3]{9y^2}$

Answer to Problem 63WE

  $3\frac{\sqrt[3]{2y}}{y}$

Explanation of Solution

Given Information:

Given expression to evaluate:

  $\sqrt[3]{6y^{-4}}\cdot \sqrt[3]{9y^2}$

Concept Used :

• Definition of nth root:

• Words : For any real numbers a and b , and any positive integer n, if $a^n=b$ then a is the nth root of b. That is $a=\sqrt[n]{b}$
• Example : Because ${(-3)}^4=81$ , -3 is a fourth root of 81 and 3 is a principal root. That is $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

    $a^{-n}=\frac{1}{a^n}$

• The non-negative root is called the Principal Root.
• Suppose n is an integer greater than 1, and a is a real number.

a	n is even	n is odd
a >0	1 unique positive and 1 unique negative real root : $\pm \sqrt[n]{a}$ ; positive root is principal root.	1 unique positive and 0 negative real root : $\sqrt[n]{a}$
a < 0	0 real roots	0 positive and 1 negative real root : $-\sqrt[n]{a}$
a=0	1 real root : $\sqrt[n]{0}=0$	1 real root : $\sqrt[n]{0}=0$

• When you find an even root of an even power and the result is an odd power , you must use the absolute value of the result to ensure that the answer is nonnegative.
• If the index is odd, there is only one real root. Therefore, there is no principal root when index is odd , and absolute value symbol are not needed.
• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$

• Rationalizing the denominator :

To eliminate radicals from a denominator or fractions from a radicand , use this process.

To rationalize a denominator, multiply the numerator and denominator by a quantity so that the radicand has an exact root.

Calculation:

• Given $\sqrt[3]{6y^{-4}}\cdot \sqrt[3]{9y^2}$
• $\begin{array}{l}\sqrt[3]{6y^{-4}}\cdot \sqrt[3]{9y^2}\\ =\sqrt[3]{6y^{-4}\cdot 9y^2}\mathrm{.............}[\text{Product property of radicals]}\\ =\sqrt[3]{3^3\cdot 2\cdot y^{2-4}}\mathrm{.............}[a^{x+y}=a^x\cdot a^y]\\ =3\sqrt[3]{\frac{2}{y^2}}\cdot \frac{\sqrt[3]{y}}{\sqrt[3]{y}}\\ =3\frac{\sqrt[3]{2y}}{y}\mathrm{....................}[\text{Product property of radicals]}\end{array}$
• So, $3\frac{\sqrt[3]{2y}}{y}$ is the solution.