$$x(3-\frac{2}{x})=\frac{3}{x}$$ then the value of $$x^{2}+\frac{1}{x^2}$$ is
Expression : $$x(3-\frac{2}{x})=\frac{3}{x}$$
=> $$3x - 2 = \frac{3}{x}$$
=> $$x - \frac{1}{x} = \frac{2}{3}$$
Squaring both sides
=> $$x^2 + \frac{1}{x^2} - 2 = \frac{4}{9}$$
=> $$x^2 + \frac{1}{x^2} = \frac{22}{9} = 2\frac{4}{9}$$
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