Answer
$\dfrac{a}{10b^{2}c}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
(100ab^3c^2)^{-\frac{1}{2}}(a^6b^{-2})^{\frac{1}{4}}
,$ use the laws of exponents.
$\bf{\text{Solution Details:}}$
Using the Power of the Product Rule of the laws of exponents, which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(100ab^3c^2)^{-\frac{1}{2}}(a^6b^{-2})^{\frac{1}{4}}
\\\\=
(10^2ab^3c^2)^{-\frac{1}{2}}(a^6b^{-2})^{\frac{1}{4}}
\\\\=
\left(10^{2\cdot\left(-\frac{1}{2}\right)}a^{-\frac{1}{2}}b^{3\cdot\left( -\frac{1}{2} \right)}c^{2\cdot\left( -\frac{1}{2} \right)} \right) \left( a^{6\cdot\frac{1}{4}}b^{-2\cdot\frac{1}{4}} \right)
\\\\=
\left(10^{-\frac{2}{2}}a^{-\frac{1}{2}}b^{-\frac{3}{2} }c^{-1} \right) \left( a^{\frac{6}{4}}b^{-\frac{2}{4}} \right)
\\\\=
\left(10^{-1}a^{-\frac{1}{2}}b^{-\frac{3}{2} }c^{-1} \right) \left( a^{\frac{3}{2}}b^{-\frac{1}{2}} \right)
.\end{array}
Using the Product Rule of the laws of exponents, which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left(10^{-1}a^{-\frac{1}{2}}b^{-\frac{3}{2} }c^{-1} \right) \left( a^{\frac{3}{2}}b^{-\frac{1}{2}} \right)
\\\\=
10^{-1}a^{-\frac{1}{2}+\frac{3}{2}}b^{-\frac{3}{2}+\left(-\frac{1}{2} \right) }c^{-1}
\\\\=
10^{-1}a^{\frac{2}{2}}b^{-\frac{4}{2}}c^{-1}
\\\\=
10^{-1}a^{1}b^{-2}c^{-1}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
10^{-1}a^{1}b^{-2}c^{-1}
\\\\=
\dfrac{a^{1}}{10^{1}b^{2}c^{1}}
\\\\=
\dfrac{a}{10b^{2}c}
.\end{array}
