If (x + frac{1}{x})^2 = 3,then (x^3 + frac{1} | vedclass

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(D) Given that $(x + \frac{1}{x})^2 = 3$.Taking the square root on both sides,we get $(x + \frac{1}{x}) = \sqrt{3}$.We know the algebraic identity $(a

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(D) Given that $(x + \frac{1}{x})^2 = 3$.Taking the square root on both sides,we get $(x + \frac{1}{x}) = \sqrt{3}$.We know the algebraic identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.Substituting $a = x$ and $b = \frac{1}{x}$:$(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x)(\frac{1}{x})(x + \frac{1}{x})$.Substituting the value $(x + \frac{1}{x}) = \sqrt{3}$:$(\sqrt{3})^3 = x^3 + \frac{1}{x^3} + 3(1)(\sqrt{3})$.$3\sqrt{3} = x^3 + \frac{1}{x^3} + 3\sqrt{3}$.Subtracting $3\sqrt{3}$ from both sides:$x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3} = 0$.

If $(x + \frac{1}{x})^2 = 3$,then $(x^3 + \frac{1}{x^3})$ is equal to

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