First, we can use the formula for the sum of cubes to factor the numerator: $$p^3-1=(p-1)(p^2+p+1)$$
Then, we can rewrite the original expression as: $$\frac{p^3-1}{p-1}=\frac{(p-1)(p^2+p+1)}{p-1}$$
The factor of $(p-1)$ cancels out, leaving us with:
Show more…