Question 189

If $$x+\frac{1}{x}=2$$ then the value of $$x^{12}+\frac{1}{x^{12}}$$ is

Solution

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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