How do you simplify #(27a^-3b^12)^(1/3) / (16a^-8b^12) ^ (1/2)#?

Question

Nathan Axel · Accepted Answer

You must remember the exponent-radical rule #x^(1/n) = root(n)(x)#

Therefore,

#(root(3)(27) xx (a^-3 xx b^12)^(1/3))/(sqrt(16) xx (a^-8 xx b^12)^(1/2))#

For the expressions in parentheses, you must calculate using the power of exponents rule, or #(a^n)^m = a^(n xx m)#

#=(3 xx a^-1 xx b^4)/(4 xx a^-4 xx b^6)#

However, we need to simplify further and write without negative exponents. This can all be done using the quotient rule #a^n/a^m = a^(n- m)#

As a shortcut to not have to use the negative exponent rule #a^-n = 1/a^n#, we must apply the quotient rule from the largest exponent. For example, in #x^2/x^4#, you would make #x^4# as n and #x^2# as m, and then you would do your subtraction. You would get #1/x^2# in this problem, which is without negative exponents, and is what we want.

#= (3 xx a^(-1 - (-4)) )/(4 xx b^(6 -4))#

#= (3a^3)/(4b^2)#

Hopefully this helps!

Eva Abramson · Answer

undefined To simplify $ \left(27a^{-3}b^{12}\right)^{\frac{1}{3}} \div \left(16a^{-8}b^{12}\right)^{\frac{1}{2}} $, first simplify each expression within the parentheses, then apply the rules of exponents and fractions. After simplifying, the expression will be $ \frac{3b^4}{2a^3} $.