Question

Find product using suitable identity $\left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$ ?

Answer 1

Hint: We need to find the product of the expression $\left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$ using a suitable identity. We solve the given question using the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to get the desired result.

Complete step by step answer:
We are given an expression and are asked to find the product of the same using a suitable identity. We will be solving the given question using the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ .
Let us understand the implementation of the above identity through an example.
Example:
Find the value of $\left( x+2 \right)\left( x-2 \right)?$
The given question is of the form $\left( a+b \right)\left( a-b \right)$
The result of the given question is obtained using an identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
In our case,
a = x
b = 2
From the above identity,
$\Rightarrow \left( x+2 \right)\left( x-2 \right)={{x}^{2}}-{{2}^{2}}$
We know that the value of ${{2}^{2}}$ is equal to 4.
Substituting the same, we get,
$\therefore \left( x+2 \right)\left( x-2 \right)={{x}^{2}}-4$
According to our question,
The given expression is $\left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$
In the above expression, the terms $\left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)$ are in the form of $\left( a+b \right)\left( a-b \right)$
The result can be obtained using an identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Here,
a = x
b = $\dfrac{1}{x}$
Applying the identity, we get,
$\Rightarrow \left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)=\left( {{x}^{2}}-{{\left( \dfrac{1}{x} \right)}^{2}} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$
Simplifying the above equation, we get,
$\Rightarrow \left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)=\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$
In the above expression, the terms $\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)$ are in the form of $\left( a+b \right)\left( a-b \right)$
The result can be obtained using an identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Here,
a = ${{x}^{2}}$
b = $\dfrac{1}{{{x}^{2}}}$
Applying the identity, we get,
$\Rightarrow \left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)=\left( {{\left( {{x}^{2}} \right)}^{2}}-{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{2}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$
Simplifying the above equation, we get,
$\Rightarrow \left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)=\left( {{x}^{4}}-\dfrac{1}{{{x}^{4}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$
In the above expression, the terms $\left( {{x}^{4}}-\dfrac{1}{{{x}^{4}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$ are in the form of $\left( a+b \right)\left( a-b \right)$
The result can be obtained using an identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Here,
a = ${{x}^{4}}$
b = $\dfrac{1}{{{x}^{4}}}$
Applying the identity, we get,
$\Rightarrow \left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)=\left( {{\left( {{x}^{4}} \right)}^{2}}-{{\left( \dfrac{1}{{{x}^{4}}} \right)}^{2}} \right)$
Simplifying the above equation, we get,
$\Rightarrow \left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)=\left( {{x}^{8}}-\dfrac{1}{{{x}^{8}}} \right)$
$\therefore$ The product of $\left( x-\dfrac{1}{x} \right)\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)$ is ${{x}^{8}}-\dfrac{1}{{{x}^{8}}}$

Note: The given question is directly formula based and any mistake in writing an identity will result in an incorrect solution. We must remember that the value of the expression ${{\left( \dfrac{1}{a} \right)}^{n}}$ is the same as the expression $\dfrac{1}{{{a}^{n}}}$ .