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If $x>1$ and $x^2+\dfrac {1}{x^2}=83$, find the value of the expression$$x^3-\dfrac {1}{x^3}$$

a) $764$

b) $750$

c) $756$

d) $760$

In this question from given I tried to approximate the value of $x$ which should just above to 9 then I tried to calculate the value of cubic expression but all options are close enough to guess. Any idea to solve it?

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First, notice that

$$\left(x-\frac{1}{x}\right)^2=x^2-2+\frac{1}{x^2}=83-2=81$$

Then, use the difference of two cubes formula:

$$x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)=9\cdot(83+1)=756$$

We take the positive root of $81$ because $x>1$.

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