Chapter 9.1, Problem 104E

To determine

To explain: whether you can rationalize the denominator of the expression $\frac{2}{4+\sqrt[3]{5}}$ by

multiplying the numerator and denominator by $4-\sqrt[3]{5}$.

Answer to Problem 104E

The required answer is: No, we cannot rationalize the denominator of the expression $\frac{2}{4+\sqrt[3]{5}}$ by

multiplying the numerator and denominator by $4-\sqrt[3]{5}$.

Explanation of Solution

Given information:

The given expression: $\frac{2}{4+\sqrt[3]{5}}$

Calculation:

  $\frac{2}{4+\sqrt[3]{5}}=\frac{2}{4+\sqrt[3]{5}}\cdot \frac{4^2-4\sqrt[3]{5}+{(\sqrt[3]{5})}^2}{4^2-4\sqrt[3]{5}+{(\sqrt[3]{5})}^2}$

(The conjugate of $4+\sqrt[3]{5}$ is $4^2-4\sqrt[3]{5}+{(\sqrt[3]{5})}^2$ )

  $\begin{array}{l}=\frac{2(16-4\sqrt[3]{5}+\sqrt[3]{25})}{(4+\sqrt[3]{5})(4^2-4\sqrt[3]{5}+{(\sqrt[3]{5})}^2)}\\ =\frac{32-8\sqrt[3]{5}+2\sqrt[3]{25}}{{(4)}^3+{(\sqrt[3]{5})}^3}\\ \end{array}$

(Since $a^3+b^3=(a+b)(a^2-ab+b^2)$ )

  $\begin{array}{l}=\frac{32-8\sqrt[3]{5}+2\sqrt[3]{25}}{64+5}\\ =\frac{32-8\sqrt[3]{5}+2\sqrt[3]{25}}{69}\end{array}$

(Simplify)