If $(xyz)^4$ is divided by $(x^{-2}y^3)^{-3}(z^{1/2})^6$, what is the simplified result for nonzero $x,y,z$?
A$\dfrac{y^{13}z}{x^2}$
B$\dfrac{x^2z}{y^{13}}$
C$x^2y^{13}z$
D$\dfrac{yz^{13}}{x^2}$
Answer & Solution
Correct answer: A. $\dfrac{y^{13}z}{x^2}$
Use the laws of indices carefully. First, $(xyz)^4=x^4y^4z^4$. Also, $(x^{-2}y^3)^{-3}=x^6y^{-9}$ and $(z^{1/2})^6=z^3$. So the expression becomes $\dfrac{x^4y^4z^4}{x^6y^{-9}z^3}=x^{4-6}y^{4-(-9)}z^{4-3}=x^{-2}y^{13}z=\dfrac{y^{13}z}{x^2}$.
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