<s>[INST] My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{4x-6}{3})=1-\frac{1}{x^{2}}(\frac{x^{2}}{3})=\frac{2}{3}$.
But according to the book the right answer is option E, i.e, $2$.  
Is the book right or me ?
 [/INST] I think that the book is wrong.
$$1-\frac{4}{3x}+\frac{2}{x^2}=\frac{3x^2}{3x^2}\cdot \frac{1-\frac{4}{3x}+\frac{2}{x^2}}{1}=\frac{3x^2-4x+6}{3x^2}=\frac{2x^2+(x^2-4x+6)}{3x^2}$$
$$=\frac{2x^2+0}{3x^2}=\frac{2}{3},x\ne 0$$
</s>