Question

Multiply and simplify by factoring then write in radical form.
3rd root of (y)^10 multiplied by 3rd root of (16y)^11

Answer 1

\[\sqrt[3]{y^{10}}*\sqrt[3]{2^{44}y^{11}}=\sqrt[3]{(y^3)^3y}*\sqrt[3]{(2^{14})^3(y^3)^3y^2}\]apply third root all perfect cubes\[=y^3\sqrt[3]{y}*2^{14}y^3\sqrt[3]{y^2}\]commute and associate factors\[=16384y^6\sqrt[3]{y*y^2}=16384y^6\sqrt[3]{y^3}\]\[=16384y^6*y=16384y^7\]

Answer 2

little correction 14*3=42
16384*y^7*2^(2/3)

Answer 3

Marina, Would you please write my original problem as you understood it because someone told me I had been writing the problems incorrectly and I want to make sure I wrote this problem correctly.

Answer 4

thanks marina, so radical form is \[1634y^7\sqrt[3]{4}\](it's the whole arithmetic thing!)

Answer 5

1. You are welcome.
2. I told you that.
3. You did pretty good job here.
4. It's possible to write in exponent form
y^(10/3)*(16y)^(11/3)

Answer 6

To Mandolino: yes
I pretty often check answers there, just to be sure
http://www.wolframalpha.com/input/?i=%2816y%29%5E%2811%2F3%29*y%5E%2810%2F3%29

Answer 7

OO. I didn't pay attention that answer need to be in radical form

Answer 8

Thanks you guys, but I think you solved the wrong problem. The second radical is not correct. Check my original problem.

Answer 9

be specific

Answer 10

the 1634 should be 1638 according to wolfram

Answer 11

is this your original?\[\sqrt[3]{y^{10}}*\sqrt[3]{(16y)^{11}}\]if so, the answer seems correct with the 1634 becoming 1638

Answer 12

Yes (*.*), but no bracket around the 16y, but otherwise yes

Answer 13

i'll go to the current post since it is shorter and will respond better