+129852 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 ]
+129852 √ [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 ]
+129852 √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 ]
