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Solution
Given $$x^2-3x +1=0 $$
or $$ x^2-3x=-1 $$
or $$ (x-3)=\frac{-1}{x}$$
or $$ (x+ \frac{1}{x})=3 $$
Now according to given question $$x + \frac{1}{x} + x^2+\frac{1}{x^2}$$
or $$x+\frac{1}{x} +(x+\frac{1}{x})^2-2$$
After putting value of $$x+\frac{1}{x}$$ in above equation , it will get reduced to 10.
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