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

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

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

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

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

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

= $a^4b^16c^4$

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