Question

How do you simplify \[{{\left[ {{\left( 2x{{y}^{3}} \right)}^{2}} \right]}^{3}}\]?

Answer 1

Hint: Assume the simplified form of the given expression as ‘E’. Write the base of E as: - \[2x{{y}^{3}}=2\times x\times {{y}^{3}}\] and apply the formula of exponents and powers given as: - \[{{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}\] and simplify the value. Use the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] for further simplification and get the answer.

Complete step by step solution:
Here, we have been provided with the expression \[{{\left[ {{\left( 2x{{y}^{3}} \right)}^{2}} \right]}^{3}}\] and we are asked to simplify it. We are going to use some formulas of the topic ‘exponents and powers’.
Now, let us assume the value of the given expression as ‘E’. So, we have,
\[\Rightarrow E={{\left[ {{\left( 2x{{y}^{3}} \right)}^{2}} \right]}^{3}}\]
We can write the base, i.e., \[2x{{y}^{3}}=2\times x\times {{y}^{3}}\], so we get,
\[\Rightarrow E={{\left[ {{\left( 2\times x\times {{y}^{3}} \right)}^{2}} \right]}^{3}}\]
Now, applying the formula: - \[{{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}\], we get,
\[\Rightarrow E={{\left[ {{2}^{2}}\times {{x}^{2}}\times {{\left( {{y}^{3}} \right)}^{2}} \right]}^{3}}\]
Using the formula \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
  & \Rightarrow E={{\left[ 4{{x}^{2}}\times {{y}^{6}} \right]}^{3}} \\
 & \Rightarrow E={{4}^{3}}\times {{\left( {{x}^{2}} \right)}^{3}}\times {{\left( {{y}^{6}} \right)}^{3}} \\
\end{align}\]
Again using the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
  & \Rightarrow E=64{{x}^{2\times 3}}\times {{y}^{6\times 3}} \\
 & \Rightarrow E=64{{x}^{6}}{{y}^{18}} \\
\end{align}\]
Hence, the above expression represents the simplified form of the given exponential expression.

Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}\], \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] and \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] because they are used everywhere. There is no easier method to solve the above question. Note that you can write \[{{\left[ {{\left( 2x{{y}^{3}} \right)}^{2}} \right]}^{3}}={{\left[ {{\left( 2x{{y}^{3}} \right)}^{3}} \right]}^{2}}\], both are the same thing. You can write \[{{\left( {{x}^{2}} \right)}^{3}}\] as \[{{x}^{2}}\times {{x}^{2}}\times {{x}^{2}}\] and apply the formula \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] to get \[{{x}^{6}}\]. Similarly, you can write \[{{\left( {{y}^{6}} \right)}^{3}}\] as \[{{y}^{6}}\times {{y}^{6}}\times {{y}^{6}}\] to get \[{{y}^{18}}\]. The answer will be the same.