## Intermediate Algebra (12th Edition)

$k^{2/3}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{k^{1/3}}{k^{2/3}\cdot k^{-1}} .$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{k^{1/3}}{k^{\frac{2}{3}+(-1)}} \\\\= \dfrac{k^{1/3}}{k^{\frac{2}{3}-1}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} k^{\frac{1}{3}-\left( \frac{2}{3}-1 \right)} \\\\= k^{\frac{1}{3}-\frac{2}{3}+1} .\end{array} To simplify the expression $\dfrac{1}{3}-\dfrac{2}{3}+1 ,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $3,3,$ and $1$ is $3$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} k^{\frac{1}{3}-\frac{2}{3}+1\cdot\frac{3}{3}} \\\\= k^{\frac{1}{3}-\frac{2}{3}+\frac{3}{3}} \\\\= k^{\frac{2}{3}} \\\\= k^{2/3} .\end{array}