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Simplify: ((a^2-b^2)^3+(b^2-c^2)^3+(c^2-...

Simplify: `((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a-b^3+(b-c)^3+(c-a)^3)`

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Putting `a^2-b^2=x,b^2-c^2=y and c^2-a^2=z`, we get
`x+y+z=(a^-b^2)+(b^2-c^2)+(c^2-a^2)=0`
`rArr x^3+y^3=3xyz " "[because x+y+z=0 rArr x^3+y^3+z^3=3xyz]`
`rArr (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3 (a^2-b^2)(b^2-c^2)(c^2-a^2)......(i)`
Again Putting `a-b =p,b-c=q and c-a=r`, we get
`(p+q+r) =(a-b)+(b-c)+(c-a)=0`
`rArr p^3+q^3+r^3=3pqr " "[because p+q+r=0 rArr p^3+q^3+r^3=3pqr]`
`rArr (a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)......(ii)`
From (i) and (ii), we get
`((a^2-b^2)+(b^2-c^2)^3+(c^2-a^2)^3)/((a-b)^3+(b-c)^3+(c-a)^3)=(3(a^2-b^2)(b^2-c^2)(c^2-a^2))/(3(a-b)(b-c)(c-a))=(a+b)(b+c)(c+a)`.
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