Question

\(\sqrt[3]{72-32\sqrt5}\cdot\sqrt{7+3\sqrt5}\)

Answer 1

\(\sqrt[3]{72-32\sqrt5}\cdot\sqrt{7+3\sqrt5}\)

\(=\sqrt[3]{27-3\cdot3^2\cdot\sqrt5+3\cdot3\cdot\left(\sqrt5\right)^2-\left(\sqrt5\right)^3}\cdot\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

\(=\sqrt[3]{\left(3-\sqrt5\right)^3}\cdot\frac{\sqrt{\left(3+\sqrt5\right)^2}}{\sqrt2}=\frac{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}{\sqrt2}=\frac{9-5}{\sqrt2}\)

\(=\frac{4}{\sqrt2}=2\sqrt2\)