What is #((3^-1a^4b^-3)^-2)/((6a^2b^-1c^-2)^2)#?
2 Answers
Explanation:
There are lots of different ways to do this, but these are the steps I followed:
Using the index law
#(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
#(3^2a^(-8-4)b^(6-(-2)))/(6^2c^(-4))=(3^2a^(-12)b^8)/(6^2c^(-4))#
Using the index law
#(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)#