## College Algebra (11th Edition)

$p-\dfrac{1}{p}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(p^{1/2}-p^{-1/2})(p^{1/2}+p^{-1/2}) ,$ use the special product on the sum and difference of like terms. Then, use the laws of exponents to simplify the resulting expression. $\bf{\text{Solution Details:}}$ Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent\begin{array}{l}\require{cancel} (p^{1/2})^2-(p^{-1/2})^2 .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} p^{\frac{1}{2}\cdot2}-p^{-\frac{1}{2}\cdot2} \\\\= p^{1}-p^{-1} \\\\= p-p^{-1} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} p-\dfrac{1}{p^1} \\\\= p-\dfrac{1}{p} .\end{array}