Solution:
Let, \(\left\lfloor x \right\rfloor = a, \left\lfloor x + \frac{1}{3} \right\rfloor = b; \hspace{1cm} a, b \in \mathbb{Z} \)
Then, \(a = b\) or \(b = a + 1\)
Case 1: If \(a = b\),
\begin{align*}
{\left\lfloor x \right\rfloor}^{3} - 7 {\left\lfloor x + \frac{1}{3} \right\rfloor} &= -13
\\ ⇒ a^3 - 7a &= -13 \\
⇒ a^3 + 13 &= 7a \end{align*}
We can prove \(a ≠ b\) by showing \(a^3 + 13 \not\equiv 7a\) (mod 2)
Case 2: If \(b = a + 1\),
\begin{align*}
{\left\lfloor x \right\rfloor}^{3} - 7 {\left\lfloor x + \frac{1}{3} \right\rfloor} &= -13 \\
⇒ a^3 - 7 (a + 1) &= -13 \\
⇒ a^3 - 7a + 6 &= 0 \\
⇒ (a-1)(a-2)(a+3) &= 0 \\
\therefore a = 1, 2, -3
\end{align*}
Properties of floor function: If \( \left\lfloor x \right\rfloor = n\); where \(n \in \mathbb{Z}\) then, \(n \le x < n + 1\)
Solving for \((a, b) = (1, 2)\),
For \(a = 1\):
\begin{align*}
\left\lfloor x \right\rfloor &= 1 \\
⇒ 1 \le x < 2 \\
\therefore x = [1, 2)
\end{align*}
For \(b = 2\):
\begin{align*}
\left\lfloor x + \frac{1}{3} \right\rfloor &= 2 \\
⇒ 2 \le (x + \frac{1}{3}) < 3 \\
⇒ 2 - \frac{1}{3} \le x < 3 - \frac{1}{3} \\
⇒ \frac{5}{3} \le x < \frac{8}{3} \\
\therefore x = [\frac{5}{3}, \frac{8}{3})
\end{align*}
Combining the conditions: \( x \in [1, 2) \cap [\frac{5}{3}, \frac{8}{3})\)
\( \therefore x \in [\frac{5}{3}, 2)\)
Similarly, for \( (2, 3), (-3, -2)\) region is \([\frac{8}{3}, 3), [\frac{-7}{3}, -2) \) respectively
The final answer is:
\( [\frac{5}{3}, 2) \cup [\frac{8}{3}, 3) \cup [\frac{-7}{3}, -2) \)