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

**Answer:**

^@198^@

**Step by Step Explanation:**

- Given, ^@ \left(x + \dfrac{ 1 }{ x } \right)^2 = 36 ^@

^@ \implies x + \dfrac{ 1 } { x } = \pm 6 ^@ - 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^@