can someone check these problem

1) $-3\sqrt[3]{-3}$ + $2\sqrt[3]{162}$ + $3\sqrt[3]{81}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2. You have a problem with the images... using LaTeX on here would be easier.

1) $-3\sqrt[6]{3} - 2\sqrt[3]{192} - \sqrt[6]{360}$

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

$2\sqrt[3]{192} = 6\sqrt[6]{3}$
This is not right

$\sqrt[6]{360} = 2\sqrt[6]{5}$
This is not right

$3\sqrt[6]{3} + 6\sqrt[6]{3} + 2\sqrt[6]{5}$
This is not right

$3\sqrt[6]{3} - 2\sqrt[6]{5}$
This is not right

3. my first question i wrote the question wrong and how do i use latex on here

4. Use the math tabs.

Click on Go Advanced and then find the button labelled 'TeX'. Select the text and press the button to see how the tabs are used.

5. when i clicked tex it shows a math tag around my text but how i write square roots with it

6. Hm... to write a 6th root of 192 for example, type in

\sqrt[6]{192}

and then wrap the math tags.

7. $$\sqrt[n]{f(x)}$$ gives $\sqrt[n]{f(x)}$

For the more general square root you can go with $$\sqrt{f(x)}$$ which is $\sqrt{f(x)}$

You can also use fractions: $$\sqrt{\dfrac{a}{b}}$$ gives $\sqrt{\dfrac{a}{b}}$

And also brackets $$\sqrt[n]{\left(\dfrac{a}{b}\right)^m}$$ gives $\sqrt[n]{\left(\dfrac{a}{b}\right)^m}$

8. i fixed my first problem can someone check it

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

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

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

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

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

= $-7\sqrt[6]{3}$ - $2\sqrt[6]{5}$

9. Where did you get that?

$6\sqrt[6]{192}= 2*3\sqrt[6]{3} = 6\sqrt[6]{3}$
This is not right.

If that was so, it means $\sqrt[6]{192} = \sqrt[6]{3}$ which is not true.

10. oh it was a typo suppose to be 2

11. Even then,

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

$2\sqrt[6]{192} = 2\sqrt[6]{3\times 64} = 4\sqrt[6]{3}$

12. oh caculated wrong