Final answer:
The real solutions for the equation p³ = 125p are p = 0, p = 5, and p = -5 after factoring and solving for p.
Step-by-step explanation:
To solve the equation p³ = 125p, we first factor out a p from both sides, which gives us p(p² - 125) = 0. We then set each factor to zero and solve for p. The solutions are:
p = 0
p² - 125 = 0
p² = 125
p = ±5√125
p = ±5
Hence, we have three real solutions: p = 0, p = 5, and p = -5.