Skip to main content

$f\left( {x + \frac{1}{x}} \right) = {x^3} + \frac{1}{{{x^3}}}$,

$f(\sqrt 3 ) = ?$

$f\left( {x + \frac{1}{x}} \right) = {x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3{x^2}\frac{1}{x} - 3x\frac{1}{{{x^2}}}$

Or, $f\left( {x + \frac{1}{x}} \right) = {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right)$

Let, $x + \frac{1}{x} = t$

So, $f(t) = {t^3} - 3t$

Or, $f(\sqrt 3 ) = 0$