Question

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

Answer 1

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}} $ .

Complete step-by-step answer:
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 $ .