If x2 - 3x + 1 = 0, then the value of x^6 + x^4 + x^2 + 1x^3 will be?

Answer: Option C

$$\begin{array}{rl}& x^2-3x+1=0\\ & \Rightarrow x^2+1=3x\\ & \Rightarrow x+\frac{1}{x}=3\\ & \Rightarrow x^3+\frac{1}{x^3}+3\times 3=27\\ & \Rightarrow x^3+\frac{1}{x^3}=18\\ & \therefore \frac{x^6+x^4+x^2+1}{x^3}\\ & =\frac{x^6}{x^3}+\frac{x^4}{x^3}+\frac{x^2}{x^3}+\frac{1}{x^3}\\ & =x^3+\frac{1}{x^3}+\frac{1}{x}+x\\ & =18+3\\ & =21\end{array}$$