Answer
$\frac{1}{x^{32}y^{40}z^{48}}$
Work Step by Step
$$(\frac{x^{4}y^{5}z^{6}}{x^{-4}y^{-5}z^{-6}})^{-4}$$
Within the parentheses, group like terms and simplify.
$$=[(\frac{x^{4}}{x^{-4}})(\frac{y^{5}}{y^{-5}})(\frac{z^{6}}{z^{-6}})]^{-4}$$
When dividing exponential expressions with the same non-zero base, subtract the exponent of the denominator from the exponent of the numerator.
$$=(x^{(4-(-4))}y^{(5-(-5))}z^{(6-(-6))})^{-4}$$
$$=(x^{(4+4)}y^{(5+5)}z^{(6+6)})^{-4}$$
$$=(x^{8}y^{10}z^{12})^{-4}$$
When a product is raised to an exponent, raise each factor to that exponent.
$$=x^{(8\times(-4))}y^{(10\times(-4))}z^{(12\times(-4))}$$
$$=x^{-32}y^{-40}z^{-48}$$
When an exponent is negative, write the expression as a fraction and move the base from the numerator to the denominator (or vise versa) making the exponent positive.
$$=\frac{1}{x^{32}y^{40}z^{48}}$$
