Chapter 7.5, Problem 32P

Simplify each expression using the properties of exponents, and then write the expression in radical form.

${(27y)}^{\frac{1}{3}}{(64y)}^{\frac{1}{3}}$

To determine

To write simply the expression ${\left(27y\right)}^{\frac{1}{3}}{\left(64y\right)}^{\frac{1}{3}}$ in radical form.

Answer to Problem 32P

  ${\left(27y\right)}^{\frac{1}{3}}{\left(64y\right)}^{\frac{1}{3}}=12\sqrt[3]{y^2}$

Explanation of Solution

Given information :

Given expression ${\left(27y\right)}^{\frac{1}{3}}{\left(64y\right)}^{\frac{1}{3}}$

  $\begin{array}{l}\text{Since}{a}^{\frac{1}{n}}=\sqrt[n]{a}\mathrm{............}(i)and{\left(a^m\right)}^n=a^{m\times n}\mathrm{........}\text{(ii)}and{\left(ab\right)}^m=a^m\times b^m\mathrm{.......}(iii)\\ a^m\times a^n=a^{m+n}\mathrm{.........}(iv)\end{array}$

Now consider ${\left(27y\right)}^{\frac{1}{3}}{\left(64y\right)}^{\frac{1}{3}}={\left(3^3\right)}^{\frac{1}{3}}{y}^{\frac{1}{3}}{\left(4^3\right)}^{\frac{1}{3}}\times y^{\frac{1}{3}}fromrule(iii)$

  $\begin{array}{l}=\left(3^{3\times \frac{1}{3}}\right)\left(4^{3\times \frac{1}{3}}\right)\left(y^{\frac{1}{3}}\right)\left(y^{\frac{1}{3}}\right)\text{from}\text{rule}(iv)\\ =3\left(4\right)y^{\frac{1}{3}+\frac{1}{3}}\text{from}\text{rule}\text{(iv)}\\ =12{\left(y^2\right)}^{\frac{1}{3}}\\ =12\sqrt[3]{y^2}\text{from}\text{rule}(i)\\ \therefore {\left(27y\right)}^{\frac{1}{3}}{\left(64y\right)}^{\frac{1}{3}}=12\sqrt[3]{y^2}\end{array}$

Hence, ${\left(27y\right)}^{\frac{1}{3}}{\left(64y\right)}^{\frac{1}{3}}=12\sqrt[3]{y^2}$