Let's determine those real numbers $p$ for which the equation

$$
x^{3}-7 x+p=0
$$

has two real roots whose difference is 1.
I. solution. Let $a$ and $a+1$ be the two roots whose difference is 1, then

$$
a^{3}-7 a+p=0 \quad \text { and } \quad(a+1)^{3}-7 a-7+p=0
$$

Subtract the two equations from each other. We get:

$$
\begin{array}{r}
3 a^{2}+3 a-6=0 \\
a^{2}+a-2=0 \\
a_{1}=1, \quad a_{2}=-2
\end{array}
$$

Since we know $a$, we can now calculate $p$:

$$
\begin{array}{lrl}
\text { 1st case: } & 1-7+p=0, & p=6 \\
\text { 2nd case: } & -8+14+p=0, & p=-6
\end{array}
$$

It can be easily verified that both values of $p$ are indeed solutions to the problem.

II. solution. If $a$ and $a+1$ are real roots, then $(x-a)(x-a-1)$ can be factored out of the polynomial $x^{3}-7 x+p$:

$$
\begin{aligned}
x^{3}-7 x+p & =(x-a)(x-a-1)(x-b)= \\
& =x^{3}-(2 a+b+1) x^{2}+\left(a^{2}+2 a b+a+b\right) x-\left(a^{2} b+a b\right)
\end{aligned}
$$

By comparing the coefficients of the powers of $x$, we get:

$$
2 a+b+1=0, \quad a^{2}+2 a b+a+b=-7, \quad-\left(a^{2} b+a b\right)=-\left(a^{2}+a\right) \cdot b=p
$$

From the first equation, we express $b$ and substitute it into the second equation:

$$
a^{2}-4 a^{2}-2 a+a-2 a-1=-7
$$

from which $-3 a^{2}-3 a=-6, a \cdot(a+1)=2$, which is a quadratic equation in $a$, with solutions $a_{1}=1$ and $a_{2}=-2$. Substituting these back into one branch

$$
b_{1}=-2 a_{1}-1=-3 \quad \text { and } \quad p_{1}=-\left(a^{2}+a\right) \cdot b=6
$$

Then the roots of the equation are $1,2,-3$. On the other branch

$$
b_{2}=-2 a_{2}-1=3, \quad p_{2}=-2 \cdot 3=-6 .
$$

Then the roots of the equation are $-2,-1,3$. Therefore, the appropriate values of $p$ are -6 and 6.