Answer

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Hint: Any number $x$ if raised to the power of a number $-y$, then this is equal to the reciprocal of $x$ raised to the power of $y$: i.e. ${{x}^{-y}}=\dfrac{1}{{{x}^{y}}}$ .

Complete step by step solution:

In this question, we need to find the value of ${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}$

We already know the property that any number $x$ if raised to the power of a number $-y$,

then this is equal to the reciprocal of $x$ raised to the power of $y$:

i.e. ${{x}^{-y}}=\dfrac{1}{{{x}^{y}}}$

Using this property, we will rewrite the given expression.

After rewriting, we will get the following:

$\dfrac{1}{{{a}^{5}}}\times {{a}^{0}}\times {{a}^{5}}$

We will now rearrange the above expression as:

$\left( \dfrac{1}{{{a}^{5}}}\times {{a}^{5}} \right)\times {{a}^{0}}$

Now, we also know the property that any number x when multiplied by its reciprocal will

give the product as 1.

i.e. $x\times \dfrac{1}{x}=1$

We will use this property on the above expression to get the following:

$\left( \dfrac{1}{{{a}^{5}}}\times {{a}^{5}} \right)\times {{a}^{0}}=1\times {{a}^{0}}={{a}^{0}}$

So, we get the following expression:

${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}={{a}^{0}}$

Now, we need to find the value of ${{a}^{0}}$

We know that any number x raised to the power of zero is equal to 1.

i.e. ${{x}^{0}}=1$ , where $x$ is any number.

Using this property, we will get the following:

${{a}^{0}}=1$

Hence, we arrive at the following final expression:

${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}=1$

So, our final answer is ${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}=1$.

Note: In this question, it is very important to identify which property to use and how to use

it. You need to rearrange the terms properly in order to see and understand the problem

clearly.

Complete step by step solution:

In this question, we need to find the value of ${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}$

We already know the property that any number $x$ if raised to the power of a number $-y$,

then this is equal to the reciprocal of $x$ raised to the power of $y$:

i.e. ${{x}^{-y}}=\dfrac{1}{{{x}^{y}}}$

Using this property, we will rewrite the given expression.

After rewriting, we will get the following:

$\dfrac{1}{{{a}^{5}}}\times {{a}^{0}}\times {{a}^{5}}$

We will now rearrange the above expression as:

$\left( \dfrac{1}{{{a}^{5}}}\times {{a}^{5}} \right)\times {{a}^{0}}$

Now, we also know the property that any number x when multiplied by its reciprocal will

give the product as 1.

i.e. $x\times \dfrac{1}{x}=1$

We will use this property on the above expression to get the following:

$\left( \dfrac{1}{{{a}^{5}}}\times {{a}^{5}} \right)\times {{a}^{0}}=1\times {{a}^{0}}={{a}^{0}}$

So, we get the following expression:

${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}={{a}^{0}}$

Now, we need to find the value of ${{a}^{0}}$

We know that any number x raised to the power of zero is equal to 1.

i.e. ${{x}^{0}}=1$ , where $x$ is any number.

Using this property, we will get the following:

${{a}^{0}}=1$

Hence, we arrive at the following final expression:

${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}=1$

So, our final answer is ${{a}^{-5}}\times {{a}^{0}}\times {{a}^{5}}=1$.

Note: In this question, it is very important to identify which property to use and how to use

it. You need to rearrange the terms properly in order to see and understand the problem

clearly.

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