### If $x > 0$ and $\left( x + \dfrac{ 1 } { x } \right)^2 = 36,$ find the value of $x^3 + \dfrac{ 1 } { x^3 }.$

$198$

Step by Step Explanation:
1. Given, $\left(x + \dfrac{ 1 }{ x } \right)^2 = 36$
$\implies x + \dfrac{ 1 } { x } = \pm 6$
2. Also, given $x > 0$
$\implies x + \dfrac{1}x = 6$
Taking the cube on both sides.
$\implies \left(x + \dfrac{1}{ x }\right)^3 = 6^3$
$\implies x^3 + \dfrac{1}{ x^3 } + 3x + \dfrac{ 3 }{ x } = 216$
$\implies x^3 + \dfrac{1}{ x^3 } + 3 \left( x + \dfrac{ 1 }{ x } \right) = 216$
$\implies x^3 + \dfrac{1}{ x^3 } + 3 \times 6 = 216$
$\implies x^3 + \dfrac{1}{ x^3 } = 198$