What is #((6x^22y^12)/(12x^7y^16))^4#?

1 Answer
EZ as pi
Apr 16, 2018

#x^60/(16y^16)#

Explanation:

#((6x^22y^12)/(12x^7y^16))^4#

Simplify inside the bracket first.

#=((cancel6x^15)/(cancel12^2y^4))^4" "larr# subtract the indices of like bases

#=(x^15/(2y^4))^4#

#=x^60/(16y^16)" "larr# raised to the power of 4

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