Question

Find the value of $\dfrac{4}{{{\left( 216 \right)}^{\dfrac{-2}{3}}}}+\dfrac{1}{{{\left( 256 \right)}^{\dfrac{-3}{4}}}}+\dfrac{2}{{{\left( 243 \right)}^{\dfrac{-1}{5}}}}$.

Answer 1

Hint: First, before proceeding for this, we must know the following properties of the exponents to get the answer as $\dfrac{1}{{{x}^{-n}}}={{x}^{n}}$. Then by using the above stated property, all the rational numbers that are in the denominator of the expression comes in the multiplication with the numerator value. Then, by using another property of the exponents which states that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$, we get the final answer.

Complete step-by-step answer:
In this question, we are supposed to find the value of a given expression by solving the properties used in the exponents.
So, before proceeding for this, we must know the following properties of the exponents to get the answer easily and appropriately:
$\dfrac{1}{{{x}^{-n}}}={{x}^{n}}$
So, by using the above stated property, all the rational numbers that are in the denominator of the expression comes in the multiplication with the numerator value.
Now, by applying the above stated property to the given question as:
$\dfrac{4}{{{\left( 216 \right)}^{\dfrac{-2}{3}}}}+\dfrac{1}{{{\left( 256 \right)}^{\dfrac{-3}{4}}}}+\dfrac{2}{{{\left( 243 \right)}^{\dfrac{-1}{5}}}}=4\times {{\left( 216 \right)}^{\dfrac{2}{3}}}+1\times {{\left( 256 \right)}^{\dfrac{3}{4}}}+2\times {{\left( 243 \right)}^{\dfrac{1}{5}}}$
Then, again using the rules of powers and exponents which states that perfect squares and cubes can be written as:
$\begin{align}
  & 216={{6}^{3}} \\
 & 243={{3}^{5}} \\
 & 256={{4}^{4}} \\
\end{align}$
So, by substituting all the values of above numbers in the expression, we get:
$4\times {{\left( 216 \right)}^{\dfrac{2}{3}}}+1\times {{\left( 256 \right)}^{\dfrac{3}{4}}}+2\times {{\left( 243 \right)}^{\dfrac{1}{5}}}=4\times {{\left( {{6}^{3}} \right)}^{\dfrac{2}{3}}}+1\times {{\left( {{4}^{4}} \right)}^{\dfrac{3}{4}}}+2\times {{\left( {{3}^{5}} \right)}^{\dfrac{1}{5}}}$
Now, by using another property of the exponents which states that:
${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$
Then, by using the above stated property, we get:
$4\times {{\left( {{6}^{3}} \right)}^{\dfrac{2}{3}}}+1\times {{\left( {{4}^{4}} \right)}^{\dfrac{3}{4}}}+2\times {{\left( {{3}^{5}} \right)}^{\dfrac{1}{5}}}=4\times {{\left( 6 \right)}^{3\times }}^{\dfrac{2}{3}}+1\times {{\left( 4 \right)}^{4\times \dfrac{3}{4}}}+2\times {{\left( 3 \right)}^{5\times \dfrac{1}{5}}}$
Now, just by simplifying the above expression to get the final answer as:
$4\times {{\left( 6 \right)}^{3\times }}^{\dfrac{2}{3}}+1\times {{\left( 4 \right)}^{4\times \dfrac{3}{4}}}+2\times {{\left( 3 \right)}^{5\times \dfrac{1}{5}}}=4\times {{6}^{2}}+1\times {{4}^{3}}+2\times {{3}^{1}}$
Then, we know the values of the following powers of the number as:
$\begin{align}
  & {{6}^{2}}=36 \\
 & {{4}^{3}}=64 \\
 & {{3}^{1}}=3 \\
\end{align}$
Now, by substituting all the values in the expression we get the final answer as:
$\begin{align}
  & 4\times {{6}^{2}}+1\times {{4}^{3}}+2\times {{3}^{1}}=4\times 36+1\times 64+2\times 3 \\
 & \Rightarrow 144+64+6=214 \\
\end{align}$
So, the value of the expression comes out to be 214.
Hence, the value of the equation $\dfrac{4}{{{\left( 216 \right)}^{\dfrac{-2}{3}}}}+\dfrac{1}{{{\left( 256 \right)}^{\dfrac{-3}{4}}}}+\dfrac{2}{{{\left( 243 \right)}^{\dfrac{-1}{5}}}}$ is 214.

Note: Now, to solve these types of questions we need to know some of the basic rules of the exponents beforehand so that we can easily proceed in these types of questions. Then, some of the basic rules are:
$\begin{align}
  & {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} \\
 & {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\
 & \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \\
\end{align}$