What is #((3^-1a^4b^-3)^-2)/((6a^2b^-1c^-2)^2)#?

2 Answers
Mar 14, 2017

#(b^8c^4)/(4a^12)#

Explanation:

There are lots of different ways to do this, but these are the steps I followed:

Using the index law #(a^m)^n=a^(mn)# you can simplify as follows:

#(3^(-1*(-2))a^(4*(-2))b^((-3)*(-2)))/(6^(1*2)a^(2*2)b^(-1*2)c^(-2*2))=(3^(2)a^(-8)b^6)/(6^2a^4b^(-2)c^(-4))#

Using the index law #a^m/a^n=a^(m-n)#, you can remove the values of #a# and #b# from the denominator (the bottom of the fraction), giving

#(3^2a^(-8-4)b^(6-(-2)))/(6^2c^(-4))=(3^2a^(-12)b^8)/(6^2c^(-4))#

Using the index law #a^(-n)=1/a^n#, and conversely #1/a^(-n)=a^n#, the next step would be to swap around the values so that they all have positive indices:

#(3^2b^8c^4)/(6^2a^12)#

Simplifying gives:

#(3^2b^8c^4)/(6^2a^12)=(9b^8c^4)/(36a^12)=(b^8c^4)/(4a^12)#

Mar 14, 2017

#(b^8c^4)/(4a^12)#

Explanation:

#((3^-1a^4b^-3)^-2)/(6a^2b^-1c^-2)^2#

#:.color(red)((a^m)^n=a^(mn)#

#:.=(3^(color(red)(-1 xx -2))a^color(red)((4 xx -2))b^color(red)(-3 xx -2))/(6^(color(red)color(red)(1 xx 2))a^color(red)(2 xx 2)b^(color(red)(-1 xx 2))c^(color(red)(-2 xx 2))#

#:.=(3^2a^-8b^6)/(6^2a^4b^-2c^-4)#

#:.=(9a^-8b^6)/(36a^4b^-2c^-4)#

#:.=(9/1 xx 1/a^8 xx b^6/1)/((36a^4)/1 xx 1/b^2 xx 1/c^4)#

#:.=((9b^6)/a^8)/((36a^4)/(b^2c^4))#

#:.=color(red)(a^m xx a^n=a^(m+n)#

#:.=(9b^6)/(a^8) xx (b^2c^4)/(36a^4)#

#:.=(cancel9^color(red)1b^8c^4)/(cancel36^color(red)4a^12)#

#:.=(b^8c^4)/(4a^12)#