Question

Simplify each Expression.
56. 3rd power of the square root of 5 * the square root of 5^3
58. 6th power of the square root of 81A^4B^8

Answer 1

\[\Large \sqrt[3]{5} \times \sqrt{5^3}\]this>

Answer 2

Yes

Answer 3

I need help on both please; I am confused and don't know what to do???

Answer 4

You can write that as:
\[\Large 5^\frac13 \times 5^\frac32\]
Taking base as common..
\[\Large 5^{\frac13 + \frac32}\]

Answer 5

5 11/6

Answer 6

is that right 5 ^\[5^ \frac{ 11 }{ 6 }\]

Answer 7

@saifoo.khan

Answer 8

Yep.

Answer 9

is it simplified

Answer 10

I think it has to be in a radical

Answer 11

Now write it in radical form:

Answer 12

what about 58, I really don't get that one???

Answer 13

Try it out.

Answer 14

well I would if I had some steps to do it

Answer 15

could you help me on half of it

Answer 16

it would mean a lot

Answer 17

Okay. start with factoring 81.

Answer 18

81 has a square root of 9

Answer 19

Since the radical have the power 6. We should look for factor which can have group of SIX factors. Like 2*2*2*2*2*2.
So we would've taken out 2. But now 81 has no such factors, so it will stay inside the radical.
Do you agree till here?

Answer 20

cause no number will multiply together to get 81

Answer 21

Exactly. So 81 will remain inside the radical.

Next A. It have power of 4. Which is smaller than 6. So it will stay outside the radical too.

Agree?

Answer 22

it also had to be in radical answer as well

Answer 23

A^4B^1 is what I got so far and the square root sign 81b^2

Answer 24

\[A^4B^1\sqrt[6]{81B^2}\]

Answer 25

NO! A will remain inside the radical. Since it has power lesser than 6.\[\Huge \sqrt[6]{81A^4B^8}\]

Answer 26

What about B you can take a pair of 6 with 2 b's leftover

Answer 27

YES! You're right. I was coming to B.

\[\Large B \sqrt[6]{81A^4B^2}\]

Answer 28

is that the simplified answer

Answer 29

yes sir.

Answer 30

@saifoo.khan Here's what wolframalpha got 3^(2/3) (A^4 b^8)^(1/6)

Answer 31

\[3^\frac{ 2 }{ 3 }\sqrt[6]{A^4B^8}\]