If $$x+\frac{1}{x}=2$$, find the value of $$(x^{2}+\frac{1}{x^{2}})(x^{3}+\frac{1}{x^{3}})$$

Expression : $$x+\frac{1}{x}=2$$

Squaring both sides, we get :

=> $$(x + \frac{1}{x})^2 = 2^2$$

=> $$x^2 + \frac{1}{x^2} + 2 = 4$$

=> $$x^2 + \frac{1}{x^2} = 2$$

Now, cubing the given expression, we get :

=> $$(x + \frac{1}{x})^3 = 2^3$$

=> $$x^3 + \frac{1}{x^3} + 3.x.\frac{1}{x}.(x + \frac{1}{x}) = 8$$

=> $$x^3 + \frac{1}{x^3} + 3*2 = 8$$

=> $$x^3 + \frac{1}{x^3} = 2$$

To find : $$(x^{2}+\frac{1}{x^{2}})(x^{3}+\frac{1}{x^{3}})$$

= 2*2 = 4