How do you simplify #((a^-2b^4c^5)/(a^-4b^-4c^3))^2#?

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Answer 1 · 2017-07-14T17:02:45

#((a^(-2)b^4c^5)/(a^(-4)b^(-4)c^3))^2=a^4b^16c^4#

Let us use the identities #a^(-m)=1/a^m#, #1/a^(-m)=a^m#, #a^mxxa^n=a^(m+n)#, #a^m/a^n=a^(m-n)# and #(a^m)^n=a^(mn)#

Hence #((a^(-2)b^4c^5)/(a^(-4)b^(-4)c^3))^2#

= #((1/a^2b^4c^5)/(1/a^4xx1/b^4c^3))^2#

= #((a^4b^4b^4c^5)/(a^2c^3))^2#

= #(a^(4-2)b^(4+4)c^(5-3))^2#

= #(a^2b^8c^2)^2#

= #a^(2xx2)b^(8xx2)c^(2xx2)#

= #a^4b^16c^4#