Question 262392
Let's break this up:


{{{3^(-2)=1/3^2=1/9}}}


{{{a^(-2)=1/a^2}}}


{{{c^(-3)=1/c^3}}}


{{{d^(-4)=1/d^4}}}


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So {{{3^(-2)a^(-2)b^5=(1/9)(1/a^2)b^5=(b^5)/(9a^2)}}}


and {{{c^(-3)d^(-4)=(1/c^3)(1/d^4)=1/(c^3d^4)}}}



Altogether, {{{(3^(-2)a^(-2)b^5)/(c^(-3)d^(-4))=((b^5)/(9a^2))/(1/(c^3d^4))}}}



Now multiply by the reciprocal of the second fraction and simplify to get the final answer of {{{(b^5c^3d^4)/(9a^2)}}}



So {{{(3^(-2)a^(-2)b^5)/(c^(-3)d^(-4))}}} simplifies to {{{(b^5c^3d^4)/(9a^2)}}} where none of the variables can be equal to zero.



In other words, {{{(3^(-2)a^(-2)b^5)/(c^(-3)d^(-4))=(b^5c^3d^4)/(9a^2)}}}