Question

How do you simplify `(\frac { 3x ^ { 2} y ^ { - 5} z } { 3^ { 2} x ^ { - 1} y z ^ { 3} } ) ^ { 2}`?

Answer 1

See a solution process below:

Explanation:

First, rewrite the term within the parenthesis as:

`((3/3^2)(x^2/x^-1)(y^-5/y)(z/z^3))^2`

Next, use this rule of exponents to rewrite the expression again:

`a = a^color(red)(1)`

`((3^color(red)(1)/3^2)(x^2/x^-1)(y^-5/y^color(red)(1))(z^color(red)(1)/z^3))^2`

Then, use this rule of exponents to simplify the `3`, `y` and `z` terms:

`x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))`

`((3^color(red)(1)/3^color(blue)(2))(x^2/x^-1)(y^color(red)(-5)/y^color(blue)(1))(z^color(red)(1)/z^color(blue)(3)))^2 =>`

`((1/3^(color(blue)(2)-color(red)(1)))(x^2/x^-1)(1/y^(color(blue)(1)-color(red)(-5)))(1/z^(color(blue)(3)-color(red)(1))))^2 =>`

`((1/3^(1))(x^2/x^-1)(1/y^(color(blue)(1)+color(red)(5)))(1/z^2))^2 =>`

`((1/3^1)(x^2/x^-1)(1/y^6)(1/z^2))^2 =>`

`((1/(3^1y^6z^2))(x^2/x^-1))^2`

Next, use this rule of exponents to simplify the `x` term:

`x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))`

`((1/(3^1y^6z^2))(x^color(red)(2)/x^color(blue)(-1)))^2 =>`

`((1/(3^1y^6z^2))(x^(color(red)(2)-color(blue)(-1))))^2 =>`

`((1/(3^1y^6z^2))(x^(color(red)(2)+color(blue)(1))))^2 =>`

`((1/(3^1y^6z^2))(x^3))^2 =>`

`(x^3/(3^1y^6z^2))^2`

Now, use this rule of exponents to complete the simplification:

`(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))`

`(x^color(red)(3)/(3^color(red)(1)y^color(red)(6)z^color(red)(2)))^color(blue)(2)`

`x^(color(red)(3) xx color(blue)(2))/(3^(color(red)(1) xx color(blue)(2))y^(color(red)(6) xx color(blue)(2))z^(color(red)(2) xx color(blue)(2))) =>`

`x^6/(3^2y^12z^4) =>`

`x^6/(9y^12z^4)`

Answer 2

`(x^6)/(9y^12z^4)`

Explanation:

First simplify what is in the brackets:

Subtract the indices of like bases, and make positive indices. where necessary:

`((color(blue)(3)color(red)(x^2)color(green)(y^-5)color(purple)(z))/(color(blue)(3^2)color(red)(x^-1)color(green)(y)color(purple)(z^3)))^2 = ((color(red)(x^2 x))/(color(blue)(3)color(green)(yy^5)color(purple)(z^2)))^2`

Now square each factor in the bracket: multiply the indices:

`((color(red)(x^3))/(color(blue)(3)color(green)(y^6)color(purple)(z^2)))^2 = (x^6)/(9y^12z^4)`