+0

+1
163
4
+2448

$$2\sqrt[6]{320}-\sqrt[6]{320}-2\sqrt[6]{384}$$

$$3\sqrt[4]{243}+2\sqrt[4]{96}-2\sqrt[4]{6}$$

If you can help me solve one of these I can try the other.

Oct 23, 2018
edited by RainbowPanda  Oct 23, 2018
edited by RainbowPanda  Oct 23, 2018

#1
+5088
+2

$$\sqrt[3]{16} = \sqrt[3]{2^4} = 2\sqrt[3]{2} \\ \\ \sqrt[3]{16}+ \sqrt[3]{5}+\sqrt[3]{2} = \\ \\ 2\sqrt[3]{2} +\sqrt[3]{5}+\sqrt[3]{2} = \\ \\ 3\sqrt[3]{2} + \sqrt[3]{5}$$

.
Oct 23, 2018
#3
+2448
0

Well then, I think I got these as well..

RainbowPanda  Oct 23, 2018
#2
+2448
0

So sorry, I posted the wrong ones! Turns out I already did those 2. Changed them

Oct 23, 2018
#4
+5088
+1

It's all the same idea.  Most likely your real problem is you just (like most people) can't see powers of small primes at a glance.

$$2\sqrt[5]{320} - \sqrt[6]{320} - 2\sqrt[5]{384} = \\ 2\sqrt[5]{2^6 \cdot 5}- \sqrt[6]{2^6\cdot 5} - 2\sqrt[5]{2^7\cdot 3}\\ 2\sqrt[5]{2^5\cdot 30}-2\sqrt[6]{5}-2\cdot 2\sqrt[5]{2^2\cdot 3} =\\ 4\sqrt[5]{30}-2\sqrt[6]{5}-4\sqrt[5]{12}$$

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Oct 23, 2018