Question

How do you simplify \[{\left( {125} \right)^{\dfrac{{ - 1}}{3}}}\] ?

Answer 1

Hint: We can see this problem is from indices and powers. This number is given as having 125 as base and \[\dfrac{{ - 1}}{3}\] as power. But since we have to simplify this we can write \[\dfrac{{ - 1}}{3}\] as a cube-root also but in the denominator of the fraction. Then the same number \[{\left( {125} \right)^{\dfrac{{ - 1}}{3}}}\] is written with positive power and then we will find its cube-root. That will be our answer.

Complete step-by-step answer:
Given that \[{\left( {125} \right)^{\dfrac{{ - 1}}{3}}}\]
This is of the form \[{a^{ - n}}\] . But we can rewrite it using the laws of indices and powers as \[ \Rightarrow {a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Thus we will apply this same on the question above.
 \[ \Rightarrow {125^{\dfrac{{ - 1}}{3}}} = \dfrac{1}{{{{125}^{\dfrac{1}{3}}}}}\]
Now this power is nothing but the cube-root of that number in base.
But 125 is the perfect cube of 5 that is \[{5^3} = 125\] . Thus we will rewrite the above expression as,
 \[ \Rightarrow {125^{\dfrac{{ - 1}}{3}}} = \dfrac{1}{{{{\left( {{5^3}} \right)}^{\dfrac{1}{3}}}}}\]
Now we will use one more law of indices as
 \[{\left( {{a^n}} \right)^m} = {a^{mn}}\]
So applying this we get,
 \[ \Rightarrow {\left( {125} \right)^{\dfrac{{ - 1}}{3}}} = \dfrac{1}{{{5^{3 \times \dfrac{1}{3}}}}}\]
Thus, cancelling 3 we get
 \[ \Rightarrow {\left( {125} \right)^{\dfrac{{ - 1}}{3}}} = \dfrac{1}{{{5^1}}}\]
We know that any number raised to the power 1 is the number itself,
 \[ \Rightarrow {\left( {125} \right)^{\dfrac{{ - 1}}{3}}} = \dfrac{1}{5}\]
This is our final answer.
So, the correct answer is “$\dfrac{1}{5}$”.

Note: These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be molded according to our convenience while solving the problem. Also note that cube-root ( to the power \[\dfrac{1}{3}\] ), square-root ( to the power \[\dfrac{1}{2}\] ) are fractions with 1 as numerator and respective root in denominator but the power in our problem is first cube root then square. Hope this will help you!