Factorize: $$(d - e)^3 + (e - f)^3 + (f - d)^3$$

$3 (d - e) (e - f) (f - d)$
1. We know that \begin{aligned} (a^3 + b^3 + c^3 - 3 a b c) = (a + b + c)(a^2 + b^2 + c^2 - a b - b c - c a) && \ldots \text{(i)} \end{aligned}
2. Let us assume \begin{aligned} (d - e) = a \\ (e - f) = b \\ \text{and } (f - d) = c \end{aligned} We have+ \begin{aligned} (d - e)^3 + (e - f)^3 + (f - d)^3 = a^3 + b^3 + c^3 \end{aligned} Also, \begin{aligned} (a + b + c) = \space & (d - e) + (e - f) + (f - d) \\ = \space & d - e + e - f + f - d \\ = \space & 0 \\ \end{aligned}
3. Substituting $(a + b + c) = 0$ in $eq \text{(i)}$, we have \begin{aligned} & (a^3 + b^3 + c^3 - 3 a b c) = (a + b + c)(a^2 + b^2 + c^2 - a b - b c - c a) \\ \implies & (a^3 + b^3 + c^3 - 3 a b c) = 0 \times (a^2 + b^2 + c^2 - a b - b c - c a) \\ \implies & (a^3 + b^3 + c^3 - 3 a b c) = 0 \\ \implies & a^3 + b^3 + c^3 = 3 a b c \\ \implies & a^3 + b^3 + c^3 = 3 (d - e) (e - f) (f - d) \end{aligned}
4. Hence, $(d - e)^3 + (e - f)^3 + (f - d)^3 = 3 (d - e) (e - f) (f - d)$