If $$x+\frac{1}{x}=2$$ then the value of $$x^{12}+\frac{1}{x^{12}}$$ is
Expression : $$x+\frac{1}{x}=2$$
Squaring both sides
=> $$x^2 + \frac{1}{x^2} + 2 = 4$$
=> $$x^2 + \frac{1}{x^2} = 2$$
Cubing both sides
=> $$x^6 + \frac{1}{x^6} + 3.x.\frac{1}{x}(x+\frac{1}{x}) = 8$$
=> $$x^6 + \frac{1}{x^6} = 8-6 = 2$$
Again, squaring both sides, we get :
=> $$x^{12} + \frac{1}{x^{12}} + 2 = 4$$
=> $$x^{12} + \frac{1}{x^{12}} = 2$$
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