# Simplify: ${{\left( 64 \right)}^{-\tfrac{2}{3}}}\times {{\left( \dfrac{1}{4} \right)}^{-3}}$ A. 4B. $\dfrac{1}{4}$ C. 1D. 16

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Hint: Recall some rules of exponents:
${{a}^{0}}=1$
${{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$
${{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}$
${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$
${{a}^{\tfrac{x}{y}}}={{\left( \sqrt[y]{a} \right)}^{x}}=\sqrt[y]{{{a}^{x}}}$
If ${{a}^{x}}=b$ , then we say that ${{b}^{\tfrac{1}{x}}}=a$ .
Observe that $64={{2}^{6}}$ and $4={{2}^{2}}$ .

We observe that $64=4\times 4\times 4$ .
The given expression ${{\left( 64 \right)}^{-\tfrac{2}{3}}}\times {{\left( \dfrac{1}{4} \right)}^{-3}}$ can be written as:
= ${{\left( {{4}^{3}} \right)}^{-\tfrac{2}{3}}}\times {{\left( \dfrac{1}{4} \right)}^{-3}}$
Using the rule ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ , we get:
= ${{4}^{3\times \left( -\tfrac{2}{3} \right)}}\times {{\left( \dfrac{1}{4} \right)}^{-3}}$
Using ${{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$ , we get:
= ${{4}^{-2}}\times \dfrac{1}{{{\left( \dfrac{1}{4} \right)}^{3}}}$
= ${{4}^{-2}}\times {{4}^{3}}$
Using ${{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}$ , we get:
= ${{4}^{-2+3}}$
= ${{4}^{1}}$
The correct answer is A. 4.

Note: Fractional powers with even denominators of negative quantities are complex numbers, and their rules of exponents are a little more exact.
Say, for instance: $\sqrt{-2}\times \sqrt{-3}\ne \sqrt{-2\times -3}$ .
${{0}^{0}}$ is not defined.
If ${{a}^{x}}\times {{a}^{y}}={{a}^{m}}\times {{a}^{n}}$ , then it is not necessary that $x=m$ and $y=n$ .