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How do you simplify: \[{{\left( 7{{x}^{2}}yz \right)}^{3}}\]?

Answer
VerifiedVerified
542.4k+ views
Hint: Assume the simplified form of the given expression as ‘E’. Write the base of E as: \[7{{x}^{2}}yz=7\times {{x}^{2}}\times y\times z\] 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 \[{{\left( 7{{x}^{2}}yz \right)}^{3}}\] 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,
\[\Rightarrow E={{\left( 7{{x}^{2}}yz \right)}^{3}}\]
We can write the base of the given exponential expression, i.e., \[7{{x}^{2}}yz\] as \[7\times {{x}^{2}}\times y\times z\], so we get,
\[\Rightarrow E={{\left( 7\times {{x}^{2}}\times y\times z \right)}^{3}}\]
Now, applying the formula: - \[{{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}\], we get,
\[\Rightarrow E={{7}^{3}}\times {{\left( {{x}^{2}} \right)}^{3}}\times {{y}^{3}}\times {{z}^{3}}\]
Since, the exponent of 7 is 3, that means we have to multiply the scalar 7 three times,
\[\begin{align}
  & \Rightarrow E=7\times 7\times 7\times {{\left( {{x}^{2}} \right)}^{3}}\times {{y}^{3}}\times {{z}^{3}} \\
 & \Rightarrow E=343\times {{\left( {{x}^{2}} \right)}^{3}}\times {{y}^{3}}\times {{z}^{3}} \\
\end{align}\]
Using the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
  & \Rightarrow E=343\times {{x}^{2\times 3}}\times {{y}^{3}}\times {{z}^{3}} \\
 & \Rightarrow E=343\times {{x}^{6}}\times {{y}^{3}}\times {{z}^{3}} \\
 & \Rightarrow E=343{{x}^{6}}{{y}^{3}}{{z}^{3}} \\
\end{align}\]
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: - \[{{a}^{m}}\times {{b}^{n}}={{a}^{m+n}}\], \[{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}\], \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] and \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] because they are used everywhere. Note that you can write: \[{{\left( {{x}^{2}} \right)}^{3}}={{\left( {{x}^{3}} \right)}^{2}}\], both are the same thing because after all the product of exponents is going to be 6. You can write \[{{\left( {{x}^{2}} \right)}^{3}}\] as \[{{x}^{2}}\times {{x}^{2}}\times {{x}^{2}}\] and apply the formula \[{{a}^{m}}\times {{b}^{n}}={{a}^{m+n}}\] to get \[{{x}^{6}}\]. The answer will be the same.
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