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

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Answer 1 · 2017-06-20T14:14:46

See a solution process below:

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 · 2017-06-20T16:10:52

$4/7$

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

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

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