$\dfrac {x^{2}} {x}+\dfrac {3x} {x}+\dfrac {1} {x}=0$
$x+3+\dfrac {1} {x}=0$
$\left( x+\dfrac {1} {x}\right) ^{2}=\left(-3\right) ^{2}$
$x^{2}+\dfrac {1} {x^{2}}+2=9$
$x^{2}+\dfrac {1} {x^{2}}=7$
$\left( x-\dfrac {1} {x}\right) ^{2}=x^{2}+\dfrac {1} {x^{2}}-2$
$\left( x-\dfrac {1} {x}\right) ^{2}=5$
$\left( x-\dfrac {1} {x}\right) =\sqrt {5} $
$x-\dfrac {1} {x}=-\sqrt {5}$
$x^{2}-\dfrac {1} {x^{2}}=\left( x-\dfrac {1} {x}\right) \left( x+\dfrac {1} {x}\right)$