Simplify Radicals: -3\sqrt[6]{3}, 2\sqrt[3]{192}, \sqrt[6]{320} and More!

[zelda1850]

im not sure if I am doing these questions correctly can someone check it

1) -3$$\sqrt[6]{3}$$ - 2$$\sqrt[3]{192}$$ - $$\sqrt[6]{320}$$

-3$$\sqrt[6]{3}$$ = -3$$\sqrt[6]{3}$$

2$$\sqrt[3]{192}$$ = 6$$\sqrt[6]{3}$$

$$\sqrt[6]{320}$$ = 2$$\sqrt[6]{5}$$

3$$\sqrt[6]{3}$$ + 6$$\sqrt[6]{3}$$ + 2$$\sqrt[6]{5}$$

= 3$$\sqrt[6]{3}$$ - 2$$\sqrt[6]{5}$$

2) -3$$\sqrt[3]{-3}$$ + 2$$\sqrt[3]{162}$$ + 3$$\sqrt[3]{81}$$

-3$$\sqrt[3]{-3}$$ = -3$$\sqrt[3]{3}$$

2$$\sqrt[3]{162}$$ = 6$$\sqrt[3]{6}$$

3$$\sqrt[3]{81}$$ = 9$$\sqrt[3]{3}$$

-3$$\sqrt[3]{3}$$ + 9$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

= 6$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

3) 4$$\sqrt[6]{3}$$ + 2$$\sqrt[4]{32}$$ - 3$$\sqrt[6]{192}$$
- 2$$\sqrt[6]{192}$$

4$$\sqrt[6]{3}$$ = 4$$\sqrt[6]{3}$$

2$$\sqrt[4]{32}$$ = 4$$\sqrt[4]{2}$$

3$$\sqrt[6]{192}$$ = 6$$\sqrt[6]{3}$$

2$$\sqrt[6]{192}$$ = 4$$\sqrt[6]{3}$$

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

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


4) $$\sqrt[-3]{320}$$ - 4$$\sqrt[4]{5}$$ + 2$$\sqrt[3]{135}$$
+ 2$$\sqrt[3]{16}$$

$$\sqrt[-3]{320}$$ = -4$$\sqrt[3]{5}$$

4$$\sqrt[4]{5}$$ = 4$$\sqrt[3]{5}$$

2$$\sqrt[3]{135}$$ = 6$$\sqrt[3]{5}$$

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

4$$\sqrt[3]{5}$$ - 4$$\sqrt[3]{5}$$ + 6$$\sqrt[3]{5}$$
+ 4$$\sqrt[3]{2}$$

= 4$$\sqrt[3]{2}$$ + 6$$\sqrt[3]{5}$$


5) 2$$\sqrt[3]{6}$$ - $$\sqrt[6]{6}$$ + 3$$\sqrt[3]{6}$$ - 3$$\sqrt[6]{384}$$

3$$\sqrt[6]{384}$$ = 2$$\sqrt[6]{6}$$

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

= 1$$\sqrt[3]{6}$$ + 5$$\sqrt[6]{6}$$

 

[eumyang]

First off, there is no need to put the numbers outside of the radical outside of the LaTex tags. You can put an entire expression into a single pair of LaTex tags, so instead of
-3$$\sqrt[6]{3}$$ - 2$$\sqrt[3]{192}$$ - $$\sqrt[6]{320}$$
you can write
$$-3\sqrt[6]{3} - 2\sqrt[3]{192} - \sqrt[6]{320}$$

Second, a lot of the radicals in your work are missing the little number in front of the radical symbol. For instance, you wrote this:
2$$\sqrt[3]{192}$$ = 4$$\sqrt{3}$$
in the 2nd line of your work in #1. Where did the little 3 go? And even after considering that fact that this is supposed to be a cube root, this is still wrong.
$$2\sqrt[3]{192} \ne 4\sqrt[3]{3}$$

2) -3$$\sqrt[3]{-3}$$ + 2$$\sqrt[3]{162}$$ + 3$$\sqrt[3]{81}$$

-3$$\sqrt[3]{-3}$$ = -3$$\sqrt[3]{-3}$$

Pull the negative outside of the radical. You can do this, because you are taking a cube root.

-3$$\sqrt[3]{-3}$$ + 6$$\sqrt{6}$$ + 9$$\sqrt{3}$$

= 6$$\sqrt[3]{3}$$ + 6$$\sqrt{6}$$

Looks like you combined the first and last radical? Can't do that -- the numbers inside are not the same.
$$-3\sqrt[3]{-3} + 9\sqrt{3} \ne 6\sqrt[3]{3}$$
However, after pulling out the negative in the first radical as I suggested earlier, you can then combine them.

Please edit your post and add the little numbers in front of the radical signs, and then we can re-check.

 

[zelda1850]

ok so i was missing the little number am i doing question 2 correctly now?

-3$$\sqrt[3]{-3}$$ + 2$$\sqrt[3]{162}$$ + 3$$\sqrt[3]{81}$$

-3$$\sqrt[3]{-3}$$ = -3$$\sqrt[3]{3}$$

2$$\sqrt[3]{162}$$ = 6$$\sqrt[3]{6}$$

3$$\sqrt[3]{81}$$ = 9$$\sqrt[3]{3}$$

-3$$\sqrt[3]{3}$$ + 9$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

= 6$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

 

[Mark44]

ok so i was missing the little number am i doing question 2 correctly now?

-3$$\sqrt[3]{-3}$$ + 2$$\sqrt[3]{162}$$ + 3$$\sqrt[3]{81}$$

-3$$\sqrt[3]{-3}$$ = -3$$\sqrt[3]{3}$$

2$$\sqrt[3]{162}$$ = 6$$\sqrt[3]{6}$$

3$$\sqrt[3]{81}$$ = 9$$\sqrt[3]{3}$$

-3$$\sqrt[3]{3}$$ + 9$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

= 6$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

No. There's an error right near the beginning. In addition, it's very difficult to follow your work for two reasons: 1) you have separate equations for each term, so it's difficult to see what your original expression is actually equal to; 2) all of the [SIZE="4'] tags really clutter up your work, making it nearly impossible to see the important stuff.

Since you are using LaTeX, just use one pair of tags for each line, as in the following.

$$-3\sqrt[3]{-3} + 2\sqrt[3]{162} + 3\sqrt[3]{81}$$
$$= 3\sqrt[3]{3} + 2\sqrt[3]{6 \cdot 3^3} + 3\sqrt[3]{3 \cdot 3^3}$$

And so on.

 

[HallsofIvy]

ok so i was missing the little number am i doing question 2 correctly now?

-3$$\sqrt[3]{-3}$$ + 2$$\sqrt[3]{162}$$ + 3$$\sqrt[3]{81}$$

-3$$\sqrt[3]{-3}$$ = -3$$\sqrt[3]{3}$$

No. $$\sqrt[3]{-3}= \sqrt[3]{(-1)(3)}= \sqrt[3]{-1}\sqrt[3]{3}= -\sqrt[3]{3}$$

2$$\sqrt[3]{162}$$ = 6$$\sqrt[3]{6}$$

3$$\sqrt[3]{81}$$ = 9$$\sqrt[3]{3}$$

These two are correct.

-3$$\sqrt[3]{3}$$ + 9$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$

= 6$$\sqrt[3]{3}$$ + 6$$\sqrt[3]{6}$$