Question

If x = 3+√8 ,find the value of x^3+1/x^3

Answer 1

Answer:

$\implies \: \boxed{ \red{ \sf {x}^{3} + \dfrac{1}{ {x}^{3} } = 198}}$

★ GivEn :-

• $\sf \: x = 3 + \sqrt{8}$

★ To Find :-

• $\sf {x}^{3} + \dfrac{1}{ {x}^{3} }$

Step-by-step explanation:

$\sf \: x = 3 + \sqrt{8}$

$\implies \sf \: \dfrac{1}{x} = \dfrac{1}{3 + \sqrt{8} }$

$\implies \sf \: \dfrac{1}{x} = \dfrac{(3 - \sqrt{8}) }{(3 + \sqrt{8}) (3 - \sqrt{8})}$

$\implies \sf \: \dfrac{1}{x} = \dfrac{(3 - \sqrt{8}) }{9 - 8}$

$\implies \sf \: \dfrac{1}{x} = {(3 - \sqrt{8}) }$

$\sf \: Now,(x + \dfrac{1}{x} ) = ( 3 + \sqrt{8} + 3 - \sqrt{8} )$

$\implies \: \sf \: (x + \dfrac{1}{x} ) = 6$

$\sf \: Now, \boxed{ \sf {x}^{3} + \dfrac{1}{ {x}^{3} } }$

$\implies \sf \: (x + \dfrac{1}{ {x} } )^{3} - 3 \times x \times \dfrac{1}{x} (x + \dfrac{1}{ {x} })$

( Putting the value..)

$\implies \sf \: {6}^{3} - 3 \times 6 = 216 - 18$

$\implies \: \boxed{ \red{ \sf {x}^{3} + \dfrac{1}{ {x}^{3} } = 198}}$