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 ?

Divide both sides by $x^2$ to get  $$1-\dfrac4x+\dfrac6{x^2}=0\iff\dfrac4x-\dfrac6{x^2}=1$$
$$1-\dfrac4{3x}+\dfrac2{x^2}=1-\dfrac13\left(\dfrac4x-\dfrac6{x^2}\right)=?$$
