Answer
$\frac{8x^{15}y^8z}{25}$
Work Step by Step
(2x$^3$y$^4$z)$^3$$\times$(5x$^{-3}$y$^2$z)$^{-2}$
=(8x$^9$y$^{12}$z$^3$)$\times$($\frac{1}{25}$x$^6$y$^{-4}$z$^{-2}$)
$=\frac{8(x^9)(x^6)(y^{12})(z^3)}{25(y^4)(z^2)}$
$=\frac{8x^{15}y^8z}{25}$
Intermediate Algebra: Connecting Concepts through Application
by Clark, Mark; Anfinson, Cynthia
Published by
Brooks Cole
ISBN 10:
0-53449-636-9
ISBN 13:
978-0-53449-636-4
Chapter 3 - Exponents, Polynomials and Functions - Chapter Review Exercises - Page 286: 1
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