If $$x^2-4x+1=0$$, then the value of $$\frac{x^6+1}{x^3}$$ is
Given : $$x^2-4x+1=0$$
=> $$x^2+1=4x$$
=> $$\frac{x^2+1}{x}=4$$
=> $$(x+\frac{1}{x})=4$$ --------------(i)
Cubing both sides, we get :
=> $$(x+\frac{1}{x})^3=(4)^3$$
=> $$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}(x+\frac{1}{x})=64$$
Substituting value from equation (i)
=> $$(x^3+\frac{1}{x^3})+3(4)=64$$
=> $$\frac{x^6+1}{x^3}=64-12=52$$
=> Ans - (B)
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