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Question

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$ \left( a \right){\text{ 76}} $

$ \left( b \right){\text{ 52}} $

$ \left( c \right){\text{ 64}} $

$ \left( d \right){\text{ None of these}} $

Answer

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Formula used:

The algebraic formula will be given by

$ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $

Here, $ a\& b $ will be the variables.

We have the equation given as $ {x^4} + \dfrac{1}{{{x^4}}} = 194 $

Now on adding both the sides $ 2 $ , we get the equation as

$ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = 194 + 2 $

Now by using the formula, the left side of the equation will be given as

$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = 196 $

By removing the square from the left sides of the equation, we will get the equation as

$ \Rightarrow \left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 14 $

Now again adding $ 2 $ both the sides of the equation, we will get

$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 14 + 2 $

Again we can see that the left side of the equation is following the formula $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $

$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = 16 $

By removing the square from the left sides of the equation, we will get the equation as

$ \Rightarrow x + \dfrac{1}{x} = 4 $ , and we will name it equation $ 1 $

Now for finding the $ {x^3} + \dfrac{1}{{{x^3}}} $ , we will take out the cube root of the equation $ 1 $ .

$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {4^3} $

We get,

\[ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x + \dfrac{1}{x}} \right)\]

And from this, we will get

$ \Rightarrow {4^3} = {x^3} + \dfrac{1}{{{x^3}}} +3 \times 4 $

And on solving and taking the constant term to one side, we get the equation as

$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 52 $

Hence, the option $ \left( b \right) $ is correct.