$$\frac{1}{n^2+7n+6}-\frac{1}{n+1} = \frac{3}{n^2+7n+6}$$

$$ \begin{aligned} \frac{1}{n^2+7n+6}-\frac{1}{n+1} &= \frac{3}{n^2+7n+6}&& \text{multiply ALL terms by } \color{blue}{ (n^2+7n+6)(n+1) }. \\[1 em](n^2+7n+6)(n+1)\cdot\frac{1}{n^2+7n+6}-(n^2+7n+6)(n+1)\cdot\frac{1}{n+1} &= (n^2+7n+6)(n+1)\cdot\frac{3}{n^2+7n+6}&& \text{cancel out the denominators} \\[1 em]n+1-(1n^2+7n+6) &= 3n+3&& \text{simplify left side} \\[1 em]n+1-n^2-7n-6 &= 3n+3&& \\[1 em]-n^2-6n-5 &= 3n+3&& \text{move all terms to the left hand side } \\[1 em]-n^2-6n-5-3n-3 &= 0&& \text{simplify left side} \\[1 em]-n^2-9n-8 &= 0&& \\[1 em] \end{aligned} $$

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