Approximate the (real) roots of $\displaystyle x^3+2x-5=0$

I used $\displaystyle x_{n+1}=x - \frac{x^3+2x-5}{3x^2+2}$

As an initial guess, I used x=1

$\displaystyle x_{n+1}= 1 - \frac{1^3+2(1)-5}{3(1)^2+2} = 1-\frac{-2}{5}=\frac{7}{5}$

I then did the itenerations out to $\displaystyle x_5$ and got the correct answer. However, when looking at the book's answer they had this:

$\displaystyle x_{n+1}=x - \frac{x^3+2x-5}{3x^2+2} = \frac{2x^3+5}{3x^2+2} $

Can someone help me see how they got from here:

$\displaystyle x_{n+1}=x - \frac{x^3+2x-5}{3x^2+2}$

to here:

$\displaystyle

\frac{2x^3+5}{3x^2+2}

$