Question

solve it guys. you guys are really intelligent​

Answer 1

Given :

It is given that $\sf x^4+\dfrac{1}{x^4}=194$

To find :

We've to find the value of $\sf x^3+\dfrac{1}{x^3}$

Solution :

$\sf : \implies {( {x}^{2} + \dfrac{1}{ {x}^{2} } )}^{2} = {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 {x}^{2} ( \dfrac{1}{ {x}^{2} } )$

$\sf : \implies( {x}^{2} + \dfrac{1}{ {x}^{2} } )^{2} = 194 + 2$

$\sf : \implies( {x}^{2} + \dfrac{1}{ {x}^{2} } )^{2} = 196$

$\sf : \implies( {x}^{2} + \dfrac{1}{ {x}^{2} } ) = \sqrt{196}$

$\sf : \implies( {x}^{2} + \dfrac{1}{ {x}^{2} } ) = 14$

Here,

$\sf : \implies(x + \dfrac{1}{x} )^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2x( \dfrac{1}{x} )$

$\sf : \implies(x + \dfrac{1}{x} )^{2} = 14 + 2$

$\sf : \implies( x + \dfrac{1}{x} )^{2} = 16$

$\sf : \implies(x + \dfrac{1}{x} ) = \sqrt{16}$

$\sf : \implies(x \times \dfrac{1}{x} ) = 4$

Now, we'll cube on both sides :

$\sf : \implies(x + \dfrac{1}{x})^{3} = {(4)}^{3}$

$\sf : \implies x^{3} + \dfrac{1}{ {x}^{3} } + 3 x( \dfrac{1}{x} (x + \dfrac{1}{x} )) = 64$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} + 3(4) = 64$

$\sf : \implies x^{3} + \dfrac{1}{ {x}^{3} } + 12 = 64$

$\sf: \implies x^{3} + \dfrac{1}{ {x}^{3} } = 64 - 12$

$\: \boxed{ \therefore \sf \: x^{3} + \dfrac{1}{ {x}^{3} } = 52}$