Question

How do you evaluate and simplify `(7^3/4^3)^(-1/3)`?

Answer 1

See a solution process below:

Explanation:

First, use this rule of exponents to eliminate the outer exponent:

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

`(7^color(red)(3)/4^color(red)(3))^color(blue)(-1/3) = 7^(color(red)(3) xx color(blue)(-1/3))/4^(color(red)(3) xx color(blue)(-1/3)) => 7^-1/4^-1`

Next, use these rules of exponents to eliminate the negative exponents:

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

`7^-1/4^-1 => 7^color(red)(-1) xx 1/4^color(red)(-1) = 1/7^color(red)(- -1) xx 4^color(red)(- -1) = 1/7^1 xx 4^1 => 4^1/7^1`

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

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

`4^color(red)(1)/7^color(red)(1) = 4/7`

Answer 2

`4/7`

Explanation:

`(7^3/4^3)^(-1/3)` is the same as `(4^3/7^3)^(+1/3)`

This is the same as `(root(3)(4^3))/(root(3)(7^3))" "=" "4/7`