+0

0
78
3
+2297

I really don't know how to do these >.<

$$-\sqrt[3]{-54n^7}$$

$$-5\sqrt[4]{128x^4}$$

$$6\sqrt[4]{567x^4}$$

$$4\sqrt[3]{64n^8}$$

Oct 22, 2018

#1
+3593
+2

$$-\sqrt[3]{-54n^7} = \\ -((-1)\sqrt[3]{54n^7} = \\ \sqrt[3]{27n^6 \cdot 2n} = \\ 3n^2\sqrt[3]{2n}$$

the rest are similar, give them another try

Oct 22, 2018
#2
+2297
+1

Thanks but I still don't understand...

RainbowPanda  Oct 22, 2018
#3
+3593
+2

The idea is to factor the stuff under the radical into two pieces.

One piece is something that has a clear cube or 4th root and the other piece is the rest.

In the first one we want to break $$-54 n^7$$ into pieces one of which we can simply take the cube root of.

$$\left(-3n^2\right)^3 = -27n^6\\ -54n^7 = -27n^6 \cdot 2n \\ \sqrt[3]{-54n^7} = \sqrt[3]{-27n^6 \cdot 2n} = -3n^2 \sqrt[3]{2n} \\ -\sqrt[3]{-54n^7} = -(-3n^2\sqrt[3]{2n}) = 3n^2\sqrt[3]{2n}$$

I'll do one more for you but I'd like to see you work these out yourself

$$-5\sqrt[4]{128x^4} = -5\sqrt[4]{2^7 x^4}=\\ -5\sqrt[4]{2^4 x^4 \cdot 2^3} = -5\cdot 2x\sqrt[4]{2^3} = \\ -10x \sqrt[4]{8}$$

Rom  Oct 22, 2018
edited by Rom  Oct 22, 2018