Question

(a²-b²)³+(b²-c²)³+(c²-a²)³/(a-b)³+(b-c)³+(c-a)³​

Answer 1

Answer:

As we know if A + B + C = 0 then A³ + B³ + C³ = 3ABC

Here in numerator

A = a² — b²

B = b² — c²

C = c² — a²

and A + B + C = a² — b² + b² — c² + c² — a² = 0

So,(a²-b²)³ + (b²-c²)³ + (c²-a²)³ = 3(a²-b²)(b²-c²)(c²-a²)

And now in denominator

A = a — b

B = b — c

C = c — a

and A + B + C = a — b + b — c + c—a = 0

So,(a-b)³ + (b-c)³ + (c-a)³ = 3(a-b)(b-c)(c-a)

Hence,

(a²-b²)³ + (b²-c²)³ + (c²-a²)³ / (a-b)³ + (b-c)³ + (c-a)³

= 3(a²-b²)(b²-c²)(c²-a²) / 3(a-b)(b-c)(c-a)

= (a+b)(b+c)(c+a)

Step-by-step explanation:

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Answer 2

Answer:

answer is

(a+b) (b+c)(c+a)