If 2x + frac{2}{x} = 1,then the value of x^{3 | vedclass

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(B) Given that $2x + \frac{2}{x} = 1$.Dividing the entire equation by $2$,we get $x + \frac{1}{x} = \frac{1}{2}$.We know the algebraic identity $(a +

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$-\frac{11}{8}$

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(B) Given that $2x + \frac{2}{x} = 1$.Dividing the entire equation by $2$,we get $x + \frac{1}{x} = \frac{1}{2}$.We know the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$.Applying this to $x + \frac{1}{x}$,we have $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x)(\frac{1}{x})(x + \frac{1}{x})$.Substituting the value $x + \frac{1}{x} = \frac{1}{2}$:$(\frac{1}{2})^{3} = x^{3} + \frac{1}{x^{3}} + 3(1)(\frac{1}{2})$.$\frac{1}{8} = x^{3} + \frac{1}{x^{3}} + \frac{3}{2}$.$x^{3} + \frac{1}{x^{3}} = \frac{1}{8} - \frac{3}{2}$.$x^{3} + \frac{1}{x^{3}} = \frac{1 - 12}{8} = -\frac{11}{8}$.

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

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