Question

Evaluate ${{\left( 0.027 \right)}^{-\dfrac{1}{3}}}$.

Answer 1

Hint: We will first open the decimal in the given term. After this we will use cube roots of 27 and 1000 by using factorization. This will help in simplifying the degree. In the last step we will reciprocal the calculated fraction and get rid of the negative power.

Complete step by step solution:
According to the question we are supposed to find out the value of the negative one third power of 0.027. For such types of evaluations, we should first solve it for the decimals. Here, in 0.027 there is the decimal up to 3 digits. So we will write 1000 in the denominator leaving 27 for the numerator. Numerically, it is written as $0.027=\dfrac{27}{1000}$…(i).
Now, we will find the cube root of 27 and 1000. If we use factorization to 27 then we will get,
$\begin{align}
  & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \,\,\,1 \\
\end{align}$
This gives us $27=3\times 3\times 3$. Since, the cube root is the number which gives its cube after multiplied to itself three times. As in this situation 3 is multiplied three times to itself so 3 is the cube root of 27.
Similarly, we get
$\begin{align}
  & 2\left| \!{\underline {\,
  1000 \,}} \right. \\
 & 2\left| \!{\underline {\,
  500 \,}} \right. \\
 & 2\left| \!{\underline {\,
  250 \,}} \right. \\
 & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \,\,\,1 \\
\end{align}$
This results in $1000=2\times 2\times 2\times 5\times 5\times 5=10\times 10\times 10$. Therefore, the cube root of 1000 is 10.
Now, we come to the term given in the question. Considering ${{\left( 0.027 \right)}^{-\dfrac{1}{3}}}$ we will substitute (i) in it. Thus, we get ${{\left( 0.027 \right)}^{-\dfrac{1}{3}}}={{\left( \dfrac{27}{1000} \right)}^{-\dfrac{1}{3}}}$.
Since cube root of 27 is 3 and 1000 is 10 so, we will write,
$\begin{align}
  & {{\left( 0.027 \right)}^{-\dfrac{1}{3}}}={{\left( \dfrac{3\times 3\times 3}{10\times 10\times 10} \right)}^{-\dfrac{1}{3}}} \\
 & \Rightarrow {{\left( 0.027 \right)}^{-\dfrac{1}{3}}}={{\left( \dfrac{{{\left( 3 \right)}^{3}}}{{{\left( 10 \right)}^{3}}} \right)}^{-\dfrac{1}{3}}} \\
 & \Rightarrow {{\left( 0.027 \right)}^{-\dfrac{1}{3}}}={{\left( {{\left( \dfrac{3}{10} \right)}^{3}} \right)}^{-\dfrac{1}{3}}} \\
 & \Rightarrow {{\left( 0.027 \right)}^{-\dfrac{1}{3}}}={{\left( \dfrac{3}{10} \right)}^{3\times -\dfrac{1}{3}}} \\
 & \Rightarrow {{\left( 0.027 \right)}^{-\dfrac{1}{3}}}={{\left( \dfrac{3}{10} \right)}^{-1}} \\
\end{align}$
To get rid of the negative sign we will simply reciprocal it. Thus, we get ${{\left( 0.027 \right)}^{-\dfrac{1}{3}}}=\dfrac{10}{3}$.
Hence, the value of ${{\left( 0.027 \right)}^{-\dfrac{1}{3}}}$ is $\dfrac{10}{3}$.

Note: Alternatively, we could solve it by first doing reciprocal of the term ${{\left( \dfrac{27}{1000} \right)}^{-\dfrac{1}{3}}}$ into ${{\left( \dfrac{1000}{27} \right)}^{\dfrac{1}{3}}}$. After this we can use the cube root method. In case, we are asked to write the answer in decimals then, we can write $\dfrac{10}{3}$ as 3.3333… . Actually, the reciprocal method that we used here is the only method and the fact to remove negative signs from the degree. Instead of using factorization we can also use a technique to figure out the cube root. For this we will consider the number 27 and look at its last digit. Since, the last digit is 7 so the cube root of it must have a last digit as 3 only. Since, after ignoring three last terms of 27 we are left with no number, so we will have 3 as a cube root of 27. Similarly, we can apply this technique to 1000. In this the last digit is 0 so its cube root will also have the last digit as 0. After ignoring the last three numbers of 1000 we are left with 1. The nearest number that divides it is 1 which is also a cube root of 1 only therefore, the cube root of 1000 is 10.