Question

How do you simplify ${( - 32)^{\dfrac{3}{5}}}?$

Answer 1

Hint:First simplify the base of the given exponential term then after simplifying it, apply the exponentials on the simplified terms and further simply it with exponential. In this problem, law of indices for multiplication, law of indices for brackets and law of indices for fractional power will be used, which are given as
${a^m} \times {a^n} = {a^{m + n}},\;{({a^m})^n} = {a^{m \times n}}\;{\text{and}}\;{a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}$ respectively.

Complete step by step answer:
In order to simplify ${( - 32)^{\dfrac{3}{5}}}$, we firstly need to simplify the base of the exponential term i.e. $ - 32$, we can write $ - 32$ as
$ - 32 = - 1 \times 2 \times 2 \times 2 \times 2 \times 2$
Using the law of indices for multiplication, we can further write it as
$
- 32 = - 1 \times {2^1} \times {2^1} \times {2^1} \times {2^1} \times {2^1} \\
\Rightarrow - 32 = - 1 \times {2^{1 + 1 + 1 + 1 + 1}} \\
\Rightarrow - 32 = - 1 \times {2^5} \\
 $
Now writing it as ${( - 32)^{\dfrac{3}{5}}}$, we will get
${( - 32)^{\dfrac{3}{5}}} = {\left( { - 1 \times {2^5}} \right)^{\dfrac{3}{5}}}$
From law of indices for brackets, we can write it further as
$
{( - 32)^{\dfrac{3}{5}}} = - {1^{\dfrac{3}{5}}} \times {2^{\dfrac{{5 \times 3}}{5}}} \\
\Rightarrow{( - 32)^{\dfrac{3}{5}}} = - {1^{\dfrac{3}{5}}} \times {2^3} \\
\Rightarrow{( - 32)^{\dfrac{3}{5}}} = - 1 \times {2^3} \\
\Rightarrow{( - 32)^{\dfrac{3}{5}}} = - {2^3} \\ $
Now using the law of indices for multiplication, we can write it as
$
{( - 32)^{\dfrac{3}{5}}} = - {2^{1 + 1 + 1}} \\
\Rightarrow{( - 32)^{\dfrac{3}{5}}} = - ({2^1} \times {2^1} \times {2^1}) \\
\Rightarrow{( - 32)^{\dfrac{3}{5}}} = - (2 \times 2 \times 2) \\
\therefore{( - 32)^{\dfrac{3}{5}}} = - 8 \\ $
So finally we have simplified ${( - 32)^{\dfrac{3}{5}}}$ and get its simplified value equals $ - 8$.

Note:We have written $ - {1^{\dfrac{3}{5}}}$ equals to $ - 1$, because three times product of negative one equals negative one and also we can write fifth root or quintic root of minus one equals to minus one itself. Let us understand this mathematically, with the help of law of indices for fractional power and brackets, $ - {1^{\dfrac{3}{5}}}$ can be written as ${\left( {\sqrt[5]{{ - 1}}} \right)^3}$.

Now we can write it as ${\left( {\sqrt[5]{{( - 1) \times ( - 1) \times ( - 1) \times ( - 1) \times ( - 1)}}} \right)^3}$ which can be with help of law of indices for multiplication, written as ${\left( {\sqrt[5]{{{{( - 1)}^5}}}} \right)^3}$ which will give ${( - 1)^3}$ by cancelling the quintic root with fifth power. Now we can write ${( - 1)^3}$ as $( - 1) \times ( - 1) \times ( - 1)$ which is equal to $( - 1)$.We can generalize it as
$ - {1^m}\;{\text{or}}\; - {1^{\dfrac{1}{m}}} = \left\{ {\begin{array}{*{20}{c}}
  { - 1}&{{\text{if}}\;{\text{m}}\;{\text{is}}\;{\text{odd}}} \\
  { - {1^m}\; = 1\;{\text{and}}\; - {1^{\dfrac{1}{m}}} = i}&{{\text{if}}\;{\text{m}}\;{\text{is}}\;{\text{even}}}
\end{array}} \right\}{\text{where}}\;m\;{\text{and}}\;i$ stands for integers and iota (Imaginary number) respectively.