Question

How do you simplify $[{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}]$ ?

Answer 1

Hint: We are given an exponential function $[{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}]$ in this question. $( - 3 \times {3^{\dfrac{1}{2}}})$ is raised to the power $ - \dfrac{1}{3}$ , so $( - 3 \times {3^{\dfrac{1}{2}}})$ is called the base and $ - \dfrac{1}{3}$ is called its power. To simplify this expression, we will do prime factorization of the base, that is, we will express the base as the product of its prime factors. Then we will try to express $( - 3 \times {3^{\dfrac{1}{2}}})$ as a number raised to some power and then we will use the law of exponent which states that when a number raised to some power is again raised to another power then keeping the base same, we multiply the powers, that is, ${({a^x})^y} = {a^{x \times y}}$ . This way we can find the simplified value of the given expression.

Complete step-by-step solution:
We have to simplify $[{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}]$
We can write the above expression as $[{( - 1 \times 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}]$
We know that $ - 1 = - 1 \times - 1 \times - 1$ and 3 is the square of $\sqrt 3 $ , so we can write –
$
  \Rightarrow [{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] = [{( - 1 \times - 1 \times - 1 \times {3^{\dfrac{1}{2}}} \times {3^{\dfrac{1}{2}}} \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] \\
   \Rightarrow [{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] = [{( - {3^{\dfrac{1}{2}}} \times - {3^{\dfrac{1}{2}}} \times - {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] \\
 $
We see that $ - {3^{\dfrac{1}{2}}}$multiplied with itself 3 times, so it can be written as –
$[{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] = {[{( - {3^{\dfrac{1}{2}}})^3}]^{\dfrac{{ - 1}}{3}}}$
We know that ${({a^x})^y} = {a^{x \times y}}$ , so we get –
$
  \Rightarrow [{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] = [{( - {3^{\dfrac{1}{2}}})^{3 \times \dfrac{{ - 1}}{3}}}] \\
   \Rightarrow [{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] = [{( - {3^{\dfrac{1}{2}}})^{ - 1}}] \\
 $
We also know that ${a^{ - x}} = \dfrac{1}{{{a^x}}}$ , so we get –
$\Rightarrow [{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}] = \dfrac{1}{{ - {3^{\dfrac{1}{2}}}}} = \dfrac{{ - 1}}{{{3^{\dfrac{1}{2}}}}} = \dfrac{{ - 1}}{{\sqrt 3 }}$
Hence, the simplified value of $[{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}]$ is $ - \dfrac{1}{{{3^{\dfrac{1}{2}}}}}$ or $ - \dfrac{1}{{\sqrt 3 }}$ .

Note: When a number is raised to the power “n”, it means that the number is multiplied with itself “n” times, for example, let ${a^n}$ be an exponential function, it means that the number “a” is multiplied with itself “n” times. So, $[{( - 3 \times {3^{\dfrac{1}{2}}})^{\dfrac{{ - 1}}{3}}}]$means that $( - 3 \times {3^{\dfrac{1}{2}}})$ is multiplied with itself $ - \dfrac{1}{3}$ times. This way we can find the value of any number that is raised to the power of some other number.