<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 your working is perfect. If the book is correct then
$$1-\frac4{3x}+\frac2{x^2}=2\implies 3x^2-4x+6=6x^2\implies3x^2+4x-6=0$$
whereas according to what is written there it must be $\;3x^2+4x-6=4x^2\;$, so I'd say the book's answer is wrong...or you forgot a coefficient $\;3\;$ for the quadratic term.
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