Question

What is the square root of 27 to the power of 3?

Answer 1

Hint: In this problem we have to find the square root of 27 which was raised to the power 3. Here the given 27 is a perfect cubic number which was given in the square root and raised to the power 3. We can then expand it as the cubic term and simplify it to get the final answer.

Complete step by step solution:
Here we have to find the square root of 27 to the power of 3
Now we can expand the problem in mathematical form, square root of 27 to the power 3.
\[\Rightarrow {{\sqrt{27}}^{3}}\]
Here we can expand the cubic term and rewrite it as
\[\Rightarrow \sqrt{27}\sqrt{27}\sqrt{27}\]
This term is in multiplication form so we can bring this all under one square root.
\[\Rightarrow \sqrt{27\times 27\times 27}\]
Now here we can make this into cubic form therefore,
\[\Rightarrow \sqrt{{{27}^{3}}}\]
Where \[{{3}^{3}}=27\] so the above term can be rewritten as
\[\Rightarrow {{\sqrt{{{\left( {{3}^{3}} \right)}^{3}}}}^{{}}}\]
We know that, the property \[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}\] we get,
\[\Rightarrow \sqrt{{{3}^{9}}}\]
Then by a property \[\sqrt{a}={{a}^{\dfrac{1}{2}}}\] , we get
\[\Rightarrow \sqrt{{{3}^{9}}}={{\left( {{3}^{9}} \right)}^{\dfrac{1}{2}}}\]
By again using the same property \[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}\] we get
\[\Rightarrow {{3}^{\dfrac{9}{2}}}\]
 We can now rewrite the power term as,
\[\Rightarrow {{3}^{4+\dfrac{1}{2}}}\]
Separating the power terms with the base by using the property \[{{a}^{b+c}}={{a}^{b}}{{a}^{c}}\] we get
\[\Rightarrow {{3}^{4}}{{3}^{\dfrac{1}{2}}}\]
By simplifying we get
\[\Rightarrow 81\times {{3}^{\dfrac{1}{2}}}\]
\[\Rightarrow 81\sqrt{3}\]
Therefore, the solution is \[81\sqrt{3}\]

Note: We should know that these types of problems are not given in mathematical form, so we have to read the given question carefully to expand the given question in mathematical form. Here we have to remember some square root properties that we have used in the problem. Students will make mistakes in expanding the cubical form and it is tough to handle when it is in the square root too.