Solve the following:

$(7^2)^3 \times [(\frac{1}{7})^2]^3$

Given :

The given expression is $(7^2)^3 \times [(\frac{1}{7})^2]^3$.

To do :

We have to solve the given expression.

Solution :

 We know that,

$a^m \times a^n = a^{m+n}, (a^m)^n = a^{m\times n}$ and $a^{-m} = \frac{1}{a^m}$

$(7^2)^3 \times [(\frac{1}{7})^2]^3 = (7)^{2 \times 3} \times (\frac{1}{7})^{2\times 3}$

                                 $= 7^6 \times (\frac{1}{7})^6$ 

                                $= 7^6 \times 7^{-6}$

                                 $= (7)^{6+(-6)}$

                                $= (7)^{6-6}$  

                                $ = 7^0$

                                $= 1$.         (Any number raised to the power of 0 is 1)

Therefore, the value of $(7^2)^3 \times [(\frac{1}{7})^2]^3$ is 1.

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Updated on: 10-Oct-2022

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