Solution
Given, $$x+\dfrac{1}{x} = 2$$
Cubing on both sides
$$(x+\dfrac{1}{x})^3 = 2^3$$
=> $$x^3+\dfrac{1}{x^3}+3\times x\times \dfrac{1}{x}(x+\dfrac{1}{x}) = 8$$
=> $$x^3+\dfrac{1}{x^3}+3(2) = 8$$
Therefore, $$x^3+\dfrac{1}{x^3} = 8-6 = 2$$
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