The value of [(a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3] ÷ [(a - b)3 + (b - c)3 + (c - a)3] is equal to: (Given a ≠ b ≠ c).
A. (a + b)(b + c)(c + a)
B. (a2 + b2)(b2 - c2)(c2 - a2)
C. (a2 + b2)(b2 + c2)(c2 + a2)
D. (a - b)(b - c)(c - a)
Answer: Option A
Solution (By Examveda Team)
[(a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3] ÷ [(a - b)3 + (b - c)3 + (c - a)3]put a = 0, b = 1, c = 2
$$\frac{{ - 1 - 27 + 64}}{{ - 1 - 1 + 8}} = \frac{{36}}{6} = 6$$
(0 + 1)(1 + 2)(2 + 0) = 3 × 2 = 6
Hence option A is right answer.
This Question Belongs to Arithmetic Ability >> Algebra
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