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#