Find all real numbers $p$ for which the equation $x^3+3px^2+(4p-1)x+p=0$ has two real roots with difference $1$.
1. Given the polynomial equation:
   \[
   x^3 + 3px^2 + (4p-1)x + p = 0
   \]
   We need to find all real numbers \( p \) for which this equation has two real roots with a difference of 1.

2. Let the roots of the polynomial be \( \alpha \), \( \beta \), and \( \gamma \). Without loss of generality, assume \( \alpha \) and \( \beta \) are the roots with a difference of 1. Thus, we can write:
   \[
   \beta = \alpha + 1
   \]
   The polynomial can be factored as:
   \[
   (x - \alpha)(x - (\alpha + 1))(x - \gamma)
   \]

3. Expanding the factored form:
   \[
   (x - \alpha)(x - \alpha - 1) = x^2 - (2\alpha + 1)x + \alpha(\alpha + 1)
   \]
   Then:
   \[
   (x^2 - (2\alpha + 1)x + \alpha(\alpha + 1))(x - \gamma)
   \]
   Expanding this product:
   \[
   x^3 - \gamma x^2 - (2\alpha + 1)x^2 + \gamma(2\alpha + 1)x + \alpha(\alpha + 1)x - \gamma \alpha(\alpha + 1)
   \]
   Simplifying:
   \[
   x^3 - (\gamma + 2\alpha + 1)x^2 + (\gamma(2\alpha + 1) + \alpha(\alpha + 1))x - \gamma \alpha(\alpha + 1)
   \]

4. Comparing coefficients with the original polynomial:
   \[
   x^3 + 3px^2 + (4p-1)x + p
   \]
   We get the following system of equations:
   \[
   \gamma + 2\alpha + 1 = -3p
   \]
   \[
   \gamma(2\alpha + 1) + \alpha(\alpha + 1) = 4p - 1
   \]
   \[
   \gamma \alpha(\alpha + 1) = -p
   \]

5. To solve these equations, we first consider the case where one of the roots is \( x = -1 \). Thus, we rewrite the polynomial as:
   \[
   x^3 + 3px^2 + (4p-1)x + p = (x + 1)(x^2 + (3p-1)x + p)
   \]

6. For the polynomial to have two roots with a difference of 1, either 0 or -2 must be a root of the quadratic polynomial:
   \[
   x^2 + (3p-1)x + p = 0
   \]

7. If \( x = 0 \) is a root:
   \[
   p = 0
   \]

8. If \( x = -2 \) is a root:
   \[
   4 - 2(3p-1) + p = 0
   \]
   Simplifying:
   \[
   4 - 6p + 2 + p = 0 \implies 6 = 5p \implies p = \frac{6}{5}
   \]

9. Alternatively, the two roots of the quadratic must have a difference of 1. The discriminant of the quadratic must be 1:
   \[
   (3p-1)^2 - 4p = 1
   \]
   Simplifying:
   \[
   9p^2 - 6p + 1 - 4p = 1 \implies 9p^2 - 10p = 0 \implies p(9p - 10) = 0
   \]
   Thus:
   \[
   p = 0 \quad \text{or} \quad p = \frac{10}{9}
   \]

10. Combining all solutions, we have:
   \[
   p = 0, \quad p = \frac{6}{5}, \quad p = \frac{10}{9}
   \]

The final answer is \(\boxed{p = 0, \quad p = \frac{6}{5}, \quad p = \frac{10}{9}}\)