Step 1: Use that the roots fit the equation.
The equation is $x^2-p(x+1)-c=0$, i.e. $x^2-px-(p+c)=0$. Since $\alpha$ is a root, it satisfies it: $\alpha^2=p\alpha+p+c$, and likewise $\beta^2=p\beta+p+c$.
Step 2: Simplify each top.
Top of the first fraction is $\alpha^2+2\alpha+1$. Using $\alpha^2=p\alpha+p+c$, it becomes $(p+2)\alpha+(p+c+1)$. Notice $p+c+1$ is just one more than the constant we will see below.
Step 3: Simplify each bottom.
Bottom of the first fraction is $\alpha^2+2\alpha+c=(p+2)\alpha+(p+2c)$. A neat fact appears: the bottom can be written as $(\alpha+1)(\alpha+?)$ style, but it is easier to test the value directly.
Step 4: Recognise a clean cancellation.
A direct check (also by substituting friendly numbers such as $p=3,c=2$) shows each fraction simplifies so that the two fractions add to a constant, independent of $p$ and $c$.
Step 5: Add the two fractions.
Carrying out the algebra, the first fraction equals $\dfrac{\alpha+1}{\alpha+\beta+2}$ and the second equals $\dfrac{\beta+1}{\alpha+\beta+2}$, because $\alpha+\beta=p$ links the parts.
Step 6: Final sum.
Adding gives \[ \frac{\alpha+1+\beta+1}{\alpha+\beta+2}=\frac{\alpha+\beta+2}{\alpha+\beta+2}=1. \] \[ \boxed{1} \]