Radical probelm can someone please check my answers?

[zelda1850]

anyone mind checking my homework and help me with the problems i got wrong

1)-5\sqrt{3} - 3\sqrt{3} = -8\sqrt{3}

2)2\sqrt{8} - \sqrt{8} = 4\sqrt{2} - 2\sqrt{2} =
2\sqrt{2}

3)-4\sqrt{6} - \sqrt{6} = -4\sqrt{6}

4)-3\sqrt{5} + 2\sqrt{5} = \sqrt{5}

5)-3\sqrt{27} - 3\sqrt{27} - 3\sqrt{27} = -9\sqrt{27}

6)-3\sqrt{12} + 3\sqrt{3} + 3\sqrt{20} = -6\sqrt{3} + 3\sqrt{3} + 6\sqrt{5}

7)-2\sqrt{45} - 3\sqrt{20} - 2\sqrt{6} = -6\sqrt{5} - 6\sqrt{5} - 2\sqrt{6} = \sqrt{5} - 2\sqrt{6}

8)\sqrt{6} * \sqrt{2} = \sqrt{12}

9)\sqrt{5} * \sqrt{3} = \sqrt{15}

10)\sqrt[3]{3} * \sqrt[3]{9} = \sqrt[3]{27} = 3

11)\sqrt[3]{-20} * \sqrt[3]{3} = \sqrt[3]{60}

 

[micromass]

3) is wrong, probably a typo...
4) is wrong: the sign is incorrect
6) is correct, but you can simplify it further: I see two \sqrt{3} there...
7) the second step is wrong, I really don't see how you got there
8) can be simplified further...
11) your sign is wrong

 

[zelda1850]

can u tell me how i can fix number 11 and number 7?

 

[micromass]

For number 11, you just need that the cube root of a negative is a negative. Thus \sqrt[3]{-20}=-\sqrt[3]{20}.

For number 7:

-6\sqrt[3]{5}-5\sqrt[3]{5}=(-6-6)\sqrt[3]{5}=-12\sqrt[3]{5}

 

[collinsmark]

Also, 5) is correct, but you can simplify it a little more too.