If $x^{2}+\frac{1}{x^{2}}=83$ and $x>1$, then the value of $x^{3}-\frac{1}{x^{3}}$ is

Use the identity: $\left(x + \dfrac{1}{x}\right)^2 = x^2 + 2 + \dfrac{1}{x^2}$

Since $x^2 + \dfrac{1}{x^2} = 83$:

$\left(x + \dfrac{1}{x}\right)^2 = 83 + 2$

$\left(x + \dfrac{1}{x}\right)^2 = 85$

$x + \dfrac{1}{x} = \pm\sqrt{85}$

Since $x > 1$, both $x$ and $\dfrac{1}{x}$ are positive, so:

$x + \dfrac{1}{x} = \sqrt{85}$

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Use the identity: $\left(x - \dfrac{1}{x}\right)^2 = x^2 - 2 + \dfrac{1}{x^2}$

$\left(x - \dfrac{1}{x}\right)^2 = 83 - 2$

$\left(x - \dfrac{1}{x}\right)^2 = 81$

$x - \dfrac{1}{x} = \pm 9$

Since $x > 1$, we have $x > \dfrac{1}{x}$, so:

$x - \dfrac{1}{x} = 9$

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Using the factorization formula:

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

$x^3 - \dfrac{1}{x^3} = 9 \times (83 + 1)$

$x^3 - \dfrac{1}{x^3} = 9 \times 84$

$x^3 - \dfrac{1}{x^3} = 756$