We want to simplify the following:
$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)}$$
First, we cancel the common factor $(x-1)$ in the left-hand fraction:
$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7}{(x+1)}$$
Now we find a common denominator to add the fractions, and see that we want both denominators to be $(x+1)^2$. To accomplish this, we can multiply the numerator and the denominator of the second fraction by the factor of $(x+1)$:
$$\frac{(7x + 6)}{(x+1)^2} - \frac{7}{(x+1)} \cdot \frac{(x+1)}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7(x+1)}{(x+1)^2}$$ $$ = \frac{(7x + 6 - 7(x+1))}{(x+1)^2} \;=\;\frac{ -1}{(x+1)^2} $$ $$\;=\;\frac{-1}{x^2 + 2x + 1}$$
You have put your finger precisely on the statement that is incorrect.
There are two competing conventions with regard to rational exponents.
The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.
In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.
The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.
The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.
Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.
A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.
But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.
Best Answer
You should get a common denominator for the last two terms by multiplying the second term by $3/3$. You have $$ \begin{align*} 1-\frac{1}{3^n}+\frac{2}{3^{n+1}} &= 1-\frac{3}{3^{n+1}}+\frac{2}{3^{n+1}}\\ &= 1-\frac{1}{3^{n+1}} \end{align*} $$