If 8x^2+9x+8=0 , then the value of x^3+1/x^3 is

Question : If $8x^2+9x+8=0$ , then the value of $x^3+\cfrac{1}{x^3}$ is :

$\cfrac{999}{212}$

$\cfrac{199}{212}$

$\cfrac{999}{512}$

$\cfrac{199}{512}$

Solution:

$8x^2+9x+8=0$
or, $8x^2+8 = -9x$
or, $8(x^2+1)=-9x$
or, $ \cfrac{x^2+1}{x} = -\cfrac{9}{8}$
or, $ \cfrac{x^2}{x} + \cfrac{1}{x} = -\cfrac{9}{8} $
or, $ x + \cfrac{1}{x} = -\cfrac{9}{8} $

$x^3 + \cfrac{1}{x^3} $
$= (x+\cfrac{1}{x})^3 -3.x.\cfrac{1}{x}(x+\cfrac{1}{x})$

$\require{cancel}$ ∴ $ x^3 + \cfrac{1}{x^3} $
$= (x+\cfrac{1}{x})^3 -3.\cancel[color:red]{x}.\cfrac{1}{\cancel[color:red]{x}}(x+\cfrac{1}{x}) $
$ = (-\cfrac{9}{8})^3 -3\times -\cfrac{9}{8} $
$ = \cfrac{-729}{512} + \cfrac{27}{8} $
$ = \cfrac{-729 + 1728}{512} $
$ = \cfrac{999}{512}$

You May Read Also :