sub and mutiply radical problem

[zelda1850]

can someone check this

1) \(\displaystyle 2\sqrt[3]{6}\) - \(\displaystyle \sqrt[6]{6}\) + \(\displaystyle 3\sqrt[3]{6}\) - \(\displaystyle 3\sqrt[6]{384}\)

\(\displaystyle 3\sqrt[6]{384}\) = \(\displaystyle 2\sqrt[6]{6}\)

\(\displaystyle 2\sqrt[3]{6}\) - \(\displaystyle \sqrt[6]{6}\) + \(\displaystyle 3\sqrt[3]{6}\) - \(\displaystyle 2\sqrt[6]{6}\)

= \(\displaystyle 2\sqrt[6]{6}\)

2) \(\displaystyle \sqrt{15}\)(\(\displaystyle 2\sqrt{10}\) - \(\displaystyle 4\sqrt{6}\))

= \(\displaystyle 10\sqrt{6}\) - \(\displaystyle 12\sqrt{10}\)

3) \(\displaystyle -3\sqrt[6]{3}\) - \(\displaystyle 2\sqrt[6]{192}\) - \(\displaystyle \sqrt[6]{320}\)

= \(\displaystyle -7\sqrt[3]{6}\) - \(\displaystyle 2\sqrt[6]{5}\)

 

[galactus]

can someone check this

1) \(\displaystyle 2\sqrt[3]{6}\) - \(\displaystyle \sqrt[6]{6}\) + \(\displaystyle 3\sqrt[3]{6}\) - \(\displaystyle 3\sqrt[6]{384}\)

\(\displaystyle 3\sqrt[6]{384}\) = \(\displaystyle 2\sqrt[6]{6}\)

\(\displaystyle 2\sqrt[3]{6}\) - \(\displaystyle \sqrt[6]{6}\) + \(\displaystyle 3\sqrt[3]{6}\) - \(\displaystyle 2\sqrt[6]{6}\)

= \(\displaystyle 2\sqrt[6]{6}\)

\(\displaystyle 5\sqrt[3]{6}-7\cdot \sqrt[6]{6}\)

2) \(\displaystyle \sqrt{15}\)(\(\displaystyle 2\sqrt{10}\) - \(\displaystyle 4\sqrt{6}\))

= \(\displaystyle 10\sqrt{6}\) - \(\displaystyle 12\sqrt{10}\)

Correct

3) \(\displaystyle -3\sqrt[6]{3}\) - \(\displaystyle 2\sqrt[6]{192}\) - \(\displaystyle \sqrt[6]{320}\)

= \(\displaystyle -7\sqrt[3]{6}\) - \(\displaystyle 2\sqrt[6]{5}\)

\(\displaystyle -2\cdot \sqrt[6]{5}-7\cdot \sqrt[6]{3}\)