Question

Find the value of ${216^{ - \dfrac{1}{{15}}}}$.

Answer 1

Hint: To solve this problem, we will first have to convert $216$ as the product of its prime factors. Then, we have to simplify the exponents of each of the prime factors and simplify the product with the exponents of the prime factors. Then, we will have to operate the product accordingly and convert the number into a simpler form. So, let us see how to solve the problem.

Complete step by step answer:
The prime factorisation of $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$.
Now, we can write it as, $216 = {2^3} \times {3^3}$.
Now, taking the exponent power, we get,
${216^{ - \dfrac{1}{{15}}}} = {\left( {{2^3} \times {3^3}} \right)^{ - \dfrac{1}{{15}}}}$
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = {2^{\left( 3 \right).\left( { - \dfrac{1}{{15}}} \right)}} \times {3^{\left( 3 \right)\left( { - \dfrac{1}{{15}}} \right)}}$
Now, simplifying the terms in power, we get,
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = {2^{\left( { - \dfrac{1}{5}} \right)}} \times {3^{\left( { - \dfrac{1}{5}} \right)}}$
We can write it as,
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = \dfrac{1}{{{2^{\dfrac{1}{5}}}}} \times \dfrac{1}{{{3^{\dfrac{1}{5}}}}}$
Now, we can write the product as,
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = {\left( {\dfrac{1}{2} \times \dfrac{1}{3}} \right)^{\dfrac{1}{5}}}$
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = {\left( {\dfrac{1}{6}} \right)^{\dfrac{1}{5}}}$
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = \dfrac{1}{{{6^{\dfrac{1}{5}}}}}$
$ \Rightarrow {216^{ - \dfrac{1}{{15}}}} = \dfrac{1}{{\sqrt[5]{6}}}$
Therefore, the value of ${216^{ - \dfrac{1}{{15}}}}$ is $\dfrac{1}{{\sqrt[5]{6}}}$.

Note: There is also an alternate way of solving this problem. We can also solve this problem by taking $216$ as the cube of $6$. We know, $216$ is the cube of $6$, so, we can directly write $216$ as ${6^3}$. So, we can then, do the rest of the problem accordingly and didn’t have to do some of the extra steps of finding the prime factors and doing their products and simplifying their exponential powers and we could have got the same answer. But doing these a bit more elaborately helps us to get more sure and it reduces the chances of silly calculation mistakes.