Nepal
Factorize:
@^
(d - e)^3 + (e - f)^3 + (f - d)^3
@^
Answer:
^@3 (d - e) (e - f) (f - d)^@
- 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}@^
- 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}@^
- 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}@^
- Hence, ^@ (d - e)^3 + (e - f)^3 + (f - d)^3 = 3 (d - e) (e - f) (f - d) ^@
