Question

Find the cube root of each of the following numbers:
${{5}^{1182}}$ .

Answer 1

Hint: In this question we are asked to find the cube root of ${{5}^{1182}}$, which is in the form of power and exponent. To find this first we need to know about the power and exponent. Then find the cubic value with the help of \[\sqrt[3]{a}={{a}^{\dfrac{1}{3}}}\].

Complete step-by-step answer:

Here, the question is given in power and exponent.
Power and exponent: we know how to calculate the expression $5\times 5$ . This expression can be written in a shorter way using something called exponent.
$5.5={{5}^{2}}$ .
An expression that represents repeated multiplication of the same factor is called power.
The number 5 is called the base, and the number 2 is called the exponent. The exponent corresponds to the number of times the base is used as a factor.
$\begin{align}
  & {{3}^{1}}=\text{3 to the power of 1 =}3 \\
 & {{3}^{2}}=\text{ 3 to the power of square =}{{\text{3}}^{2}} \\
 & {{3}^{3}}=\text{ 3 to the power of cube =}{{\text{3}}^{3}} \\
 & {{3}^{8}}=\text{ 3 to the power of 8 =}{{\text{3}}^{8}} \\
\end{align}$
For any real number ‘a’ ,
It is said that the cube root of \[\sqrt[3]{a}={{a}^{\dfrac{1}{3}}}\]
So, if a is written as power $a={{b}^{c}}$ ,
Then to calculate the cube root you have to divide the power of exponent by 3.
$\sqrt[3]{{{b}^{c}}}={{\left( {{b}^{c}} \right)}^{\dfrac{1}{3}}}={{b}^{\dfrac{1}{3}c}}={{b}^{\dfrac{c}{3}}}$ .
Here, to find the cube root of ${{5}^{1182}}$, we need 5 as ‘a’ and 1182 as ‘c’.
Now, we will substitute it in the above equation, we get –
$\begin{align}
  & \sqrt[3]{{{5}^{1182}}}={{\left( {{5}^{1182}} \right)}^{\dfrac{1}{3}}} \\
 & ={{5}^{\dfrac{1}{3}1182}} \\
 & ={{5}^{394}} \\
\end{align}$
Hence, the cube root of ${{5}^{1182}}$is ${{5}^{394}}$ .

Note: Here, students may get confused while finding the cubic value of a real number with an exponent. They should be familiar with the law of exponents’ identities and radicals. A radical is simply a fractional exponent: the square of x is just ${{x}^{\dfrac{1}{2}}}$ and the cube of x is just ${{x}^{\dfrac{1}{3}}}$ .