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:
  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^@

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