What is the value of \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ A. \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}\]?
A. \[{{a}^{16}}\]
B. \[{{a}^{12}}\]
C. \[{{a}^{8}}\]
D. \[{{a}^{4}}\]
Answer
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509.1k+ views
Hint: Take each term separately and simplify the expression by multiplying the powers. Then multiply the resultants of both terms to get the required value.
“Complete step-by-step answer:”
We are given the expression, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}\].
Now let us take the first expression, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}\].
\[\sqrt[6]{{{a}^{9}}}\]can be written as \[{{\left( {{a}^{9}} \right)}^{\dfrac{1}{6}}}\].
\[\sqrt[3]{\sqrt[6]{{{a}^{9}}}}\]can be written as \[{{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{6}}} \right)}^{\dfrac{1}{3}}}\].
Hence, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}={{\left[ {{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{6}}} \right)}^{\dfrac{1}{3}}} \right]}^{4}}\].
Now, let us multiply all the powers, and simplify it we get,
\[{{a}^{9\times \dfrac{1}{6}\times \dfrac{1}{3}\times 4}}={{a}^{\dfrac{9\times 4}{6\times 3}}}={{a}^{2}}\]
Hence, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\].
Similarly, take the other expression \[{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}\].
\[\begin{align}
& \sqrt[3]{{{a}^{9}}}={{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \\
& \sqrt[6]{{{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}}}={{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \right)}^{\dfrac{1}{6}}} \\
& \therefore {{\left[ {{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \right)}^{\dfrac{1}{6}}} \right]}^{4}}={{\left( {{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \right)}^{\dfrac{1}{6}}} \right)}^{4}} \\
\end{align}\]
Now let us multiply all the powers and simplify it,
\[{{a}^{9\times \dfrac{1}{6}\times \dfrac{1}{3}\times 4}}={{a}^{\dfrac{9\times 4}{6\times 3}}}={{a}^{2}}\]
\[\therefore {{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\]and \[{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\].
\[\therefore {{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\times {{a}^{2}}={{a}^{4}}\]
Hence, we got the value of \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\times {{a}^{2}}={{a}^{4}}\].
\[\therefore \]Option (d) is the correct answer.
Note: In a question like this, the root can be expressed as an exponential value. Don’t immediately try to take the root, but convert it to exponential form and multiply the powers. We get the value more easily. Be careful when you multiply the powers.
“Complete step-by-step answer:”
We are given the expression, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}\].
Now let us take the first expression, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}\].
\[\sqrt[6]{{{a}^{9}}}\]can be written as \[{{\left( {{a}^{9}} \right)}^{\dfrac{1}{6}}}\].
\[\sqrt[3]{\sqrt[6]{{{a}^{9}}}}\]can be written as \[{{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{6}}} \right)}^{\dfrac{1}{3}}}\].
Hence, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}={{\left[ {{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{6}}} \right)}^{\dfrac{1}{3}}} \right]}^{4}}\].
Now, let us multiply all the powers, and simplify it we get,
\[{{a}^{9\times \dfrac{1}{6}\times \dfrac{1}{3}\times 4}}={{a}^{\dfrac{9\times 4}{6\times 3}}}={{a}^{2}}\]
Hence, \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\].
Similarly, take the other expression \[{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}\].
\[\begin{align}
& \sqrt[3]{{{a}^{9}}}={{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \\
& \sqrt[6]{{{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}}}={{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \right)}^{\dfrac{1}{6}}} \\
& \therefore {{\left[ {{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \right)}^{\dfrac{1}{6}}} \right]}^{4}}={{\left( {{\left( {{\left( {{a}^{9}} \right)}^{\dfrac{1}{3}}} \right)}^{\dfrac{1}{6}}} \right)}^{4}} \\
\end{align}\]
Now let us multiply all the powers and simplify it,
\[{{a}^{9\times \dfrac{1}{6}\times \dfrac{1}{3}\times 4}}={{a}^{\dfrac{9\times 4}{6\times 3}}}={{a}^{2}}\]
\[\therefore {{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\]and \[{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\].
\[\therefore {{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\times {{a}^{2}}={{a}^{4}}\]
Hence, we got the value of \[{{\left[ \sqrt[3]{\sqrt[6]{{{a}^{9}}}} \right]}^{4}}{{\left[ \sqrt[6]{\sqrt[3]{{{a}^{9}}}} \right]}^{4}}={{a}^{2}}\times {{a}^{2}}={{a}^{4}}\].
\[\therefore \]Option (d) is the correct answer.
Note: In a question like this, the root can be expressed as an exponential value. Don’t immediately try to take the root, but convert it to exponential form and multiply the powers. We get the value more easily. Be careful when you multiply the powers.
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