Question

Hi please help me solve this problem

Answer 1

Answer:

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

Step-by-step explanation:

There are two things to make use of here.

First, in both the numerator and the denominator, the sum of the terms that get cubed is zero.  That is,

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

Second, when x+y+z=0, we have x³+y³+z³ = 3xyz.  To see this, just consider the expansion

 (x+y+z)³ = (x³+y³+z³) + 3(x+y+z)(xy+yz+zx) - 3xyz

and substitute x+y+z=0.

Putting these two facts together, the numerator is equal to

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

and the denominator is equal to

 3(a-b)(b-c)(c-a),

so the whole expression simplifies to

 (a+b)(b+c)(c+a).

Hope that helps!