## Precalculus (6th Edition)

$\color{blue}{\dfrac{\sqrt[4]{7}+\sqrt[4]{9}t^2}{t^3}}$
Simplify each radical term to obtain: $=\sqrt[4]{\dfrac{7}{(t^3)^4}}+\sqrt[4]{\dfrac{9}{t^4}} \\=\dfrac{\sqrt[4]{7}}{t^3}+\dfrac{\sqrt[4]{9}}{t}$ Make the terms similar using their LCD of $t^3$ to obtain: $=\dfrac{\sqrt[4]{7}}{t^3}+\dfrac{\sqrt[4]{9}(t^2)}{t(t^2)} \\=\dfrac{\sqrt[4]{7}}{t^3}+ \dfrac{\sqrt[4]{9}t^2}{t^3}$ Add the numerators and retain the denominator to obtain: $=\color{blue}{\dfrac{\sqrt[4]{7}+\sqrt[4]{9}t^2}{t^3}}$