Chapter 9 - Solving Quadratic Equations - 9.1 - Properties of Radicals - Exercises - Page 487: 98

$\sqrt[5] {2} \sqrt[4] {7}+2$

$\sqrt[5] {2}(\sqrt[4] {7}+\sqrt[5] {16})=\sqrt[5] {2}\cdot \sqrt[4] {7}+\sqrt[5] {2}\cdot\sqrt[5] {16}$ $=\sqrt[5] {2} \sqrt[4] {7}+\sqrt[5] {2\cdot16}$ $=\sqrt[5] {2} \sqrt[4] {7}+\sqrt[5] {32}$ $=\sqrt[5] {2} \sqrt[4] {7}+\sqrt[5] {2^{5}}$ $=\sqrt[5] {2} \sqrt[4] {7}+2$