Question 9

# If $$x^2-4x+1=0$$, then the value of $$\frac{x^6+1}{x^3}$$ is

Solution

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)