Question 577760
Note that


*[tex \LARGE x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 4]


*[tex \LARGE (x + \frac{1}{x})^2 = 6 \Rightarrow x + \frac{1}{x} = \pm \sqrt{6}]


Consider the equation


*[tex \LARGE (x^2 + \frac{1}{x^2})(x + \frac{1}{x}) = \pm 4 \sqrt{6}]


*[tex \LARGE x^3 + \frac{1}{x^3} + x + \frac{1}{x} = \pm 4 \sqrt{6}]


We know that if x + 1/x is sqrt(6), then we will have


*[tex \LARGE x^3 + \frac{1}{x^3} + \sqrt{6} = 4 \sqrt{6} \Rightarrow x^3 + \frac{1}{x^3} = 3 \sqrt{6}]


If x + 1/x = -sqrt(6) then we have


*[tex \LARGE x^3 + \frac{1}{x^3} - \sqrt{6} = -4\sqrt{6} \Rightarrow x^3 + \frac{1}{x^3} = -3\sqrt{6}]


Either of these can be correct.