Question

Simplify.
Don't give inappropriate answers otherwise...................
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Answer 1

First, we can observe that each bracket adds up to be zero.

(a-b)+(b-c)+(c-a)=0, and (a²-b²)+(b²-c²)+(c²-a²)=0

Now there is an identity that is used when the sum is zero.

→ $\dfrac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3+(b-c)^3+(c-a)^3}$

→ $\dfrac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{3(a-b)(b-c)(c-a)}$

→ $\dfrac{(a+b)(b+c)(c+a)}{1}$

So the answer is the multiplication of a+b, b+c, and c+a.

Answer 2

First, we can observe that each bracket adds up to be zero.

(a-b)+(b-c)+(c-a)=0, and (a²-b²)+(b²-c²)+(c²-a²)=0

Now there is an identity that is used when the sum is zero.

→ \dfrac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3+(b-c)^3+(c-a)^3}

(a−b)

3

+(b−c)

3

+(c−a)

3

(a

2

−b

2

)

3

+(b

2

−c

2

)

3

+(c

2

−a

2

)

3

→ \dfrac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{3(a-b)(b-c)(c-a)}

3(a−b)(b−c)(c−a)

3(a

2

−b

2

)(b

2

−c

2

)(c

2

−a

2

)

→ \dfrac{(a+b)(b+c)(c+a)}{1}

1

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

So the answer is the multiplication of a+b, b+c, and c+a.

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