If $8x^2 + 9x + 8 = 0$, then the value of $x^3 + \frac{1}{x^3}$ is :
Answer & explanation
Correct answer: option 3
If x + \(\frac{1}{x}\) = n
then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n
If $8x^2 + 9x + 8 = 0$
Now divide by 8x on both sides of the equation,
x + \(\frac{1}{x}\) = \(\frac{-9}{8}\)
then, $x^3 +\frac{1}{x^3}$ = ( \(\frac{-9}{8}\) )3 - 3 × \(\frac{9}{8}\)
$x^3 +\frac{1}{x^3}$ = \(\frac{-729}{512}\) - \(\frac{27}{8}\)
$x^3 +\frac{1}{x^3}$ = \(\frac{-729 + 1728}{512}\)
$x^3 +\frac{1}{x^3}$ = $\frac{999}{512}$