# Simplify and use positive exponent (125 X^1/2 Y^-1/3)^-2/3 (X^1/3 Y-2/3)1/2 is power on x -1/3 is power on y -2/3 is power on whole braket both brackets will multiply 1/3 is power on x -2/3 is...

Simplify and use positive exponent

(125 X^1/2 Y^-1/3)^-2/3 (X^1/3 Y-2/3)

1/2 is power on x

-1/3 is power on y

-2/3 is power on whole braket

both brackets will multiply

1/3 is power on x

-2/3 is power on y

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(125 X^1/2 Y^-1/3)^-2/3 *(X^1/3 Y-2/3)

(125x^1/2 /y^1/3 )^-2/3*(x^1/3/ y^2/3)

==> [(y^1/3)/(5^3*x^1/2)^2/3] [(x^1/3)/y^2/3]

==> [ (y^2/9)/(5^2*x^1/3)]* (x^1/3)/y^2/3

==> (y^2/9)(x^1/3)/ (25x^1/3)(y^2/3)

==>[ y^(2/9-2/3)](x^1/3-1/3)/ 25

==> (y^-4/9)(1)/ 25

==> 1/ (25y^4/9)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To simplify and use  positive power:{[125x^(1/2)y^(-1/3)]^(-2/3) } {x^(1/3)*y^(-2/3)}

Solution:

First bracket:

{125x^(1/2) y^(-1/3) }^(-2/3)

= 125^(-2/3) *x^(1/2*(-2/3)) *y((-1/3*(-2/3).

= (5^3)^(-2/3) * x^(-1/3)*y(2/9), as a^m)^n = a^(mn)

=5^(-2) * x^ (-1/3) * y^(2/9).............(1)

2nd bracket : x^(1/3)*y^(-2/3)..........(2)

So the given expression = {5^(-2) *x^(-1/3)*y^(2/9)}*{x^(1/3)*y^(-2/3)}

We can rearrange x's together and y 's together as multiplication is commutative and the expression is

5^(-2)* (x^(-1/3)*x^(1/3)) (y^2/3)*y^(-2/3))

= (1/5^2). x ^(1/3 - 1/3)* y^(2/9-6/9)

= 1/25 x^0*y^(-4/9) .

=1/(5^2* y^(4/9), as a^-m = 1/a^m.