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#1**+2 **

I find it easiest to put these in expnential form and then simplify

√ 27x^11y^6 =

(3^3)^(1/2) * (x^11)^(1/2) * (y^6)^(1/2) =

(3)^(3/2) * (x)^11/2 * (y)^3

Now.....we can find out how many powers of each thing come out of the radical and how many powers stay in by this procedure :

Divide 3/2 and express as a mixed number = 1 + 1/2

The whole number tells how many powers of 3 come out of the radical = 3^1

The numerator tells how many powers of 3 stay in the radical = 3^1

Similarly

11/2 = 5 + 1/2

5 powers of x come out......1 power stays in

And

3 = 3

We have no mixed number.....so 3 powers of y come out and there remain 0 that stay in

So....we have

3^1* x^5 * y^3 √ [ 3^1 * x^1 ] =

3x^5y^3 √ [ 3x ]

CPhill Feb 26, 2019

#2**+2 **

√ [8x^3y^8 ] =

√[ 2^3 * x^3 * y^8] =

(2^3)^(1/2) * (x^3)^(1/2) * (y^8)^(1/2) =

(2)^(3/2) * (x)^(3/2) * y^4

As before

3/2 = 1 + 1/2 1 power of 2 comes out 1 stays in

3/2 = 1 + 1/2 1 power of x comes out, 1 stays in

4 = 4 4 powers of y come out......0 remain in

So we have

2xy^4√ [ 2x ]

CPhill Feb 26, 2019

#3**+2 **

√72x^9y^4 = √2 * √36 *x^9 * y^4

We can write

√2* (36)^(1/2) * (x^9)^(1/2) * (y^4)^(1/2 ) =

√2 * 6 * x^(9/2) * (y)^(2)

The only thing here is to find out how many x's stay in and how many come out

9/2 = 4 + 1/2 4 powers come out and 1 power stays in......so we have

6x^4y^2 √ [ 2 x ]

CPhill Feb 26, 2019