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

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(C) Given: $2x + \frac{2}{x} = 3$Divide the entire equation by $2$:$x + \frac{1}{x} = \frac{3}{2}$Now,cube both sides using the identity $(a+b)^3 = a^

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$\frac{7}{8}$

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(C) Given: $2x + \frac{2}{x} = 3$Divide the entire equation by $2$:$x + \frac{1}{x} = \frac{3}{2}$Now,cube both sides using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$:$(x + \frac{1}{x})^3 = (\frac{3}{2})^3$$x^3 + \frac{1}{x^3} + 3(x)(\frac{1}{x})(x + \frac{1}{x}) = \frac{27}{8}$Substitute $x + \frac{1}{x} = \frac{3}{2}$ into the equation:$x^3 + \frac{1}{x^3} + 3(1)(\frac{3}{2}) = \frac{27}{8}$$x^3 + \frac{1}{x^3} + \frac{9}{2} = \frac{27}{8}$$x^3 + \frac{1}{x^3} = \frac{27}{8} - \frac{9}{2} = \frac{27 - 36}{8} = -\frac{9}{8}$Finally,calculate the value of $x^3 + \frac{1}{x^3} + 2$:$-\frac{9}{8} + 2 = \frac{-9 + 16}{8} = \frac{7}{8}$

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

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