If $$x^{2} - 8x + 1 = 0$$, then what is the value of $$x^{2}+\frac{1}{x^{2}}$$?
Given : $$x^2-8x+1=0$$
Dividing both sides by $$'x'$$
=> $$x+\frac{1}{x}=8$$
Squaring both sides, we get :
=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=64$$
=> $$x^2+\frac{1}{x^2}=64-2=62$$
=> Ans - (D)
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