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Taking a²=x,b²=y and c²=z
Identity,
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
if x+y+z=0
then,
x³+y³+z³=3xyz
so,
In the question,
(a²-b²)+(b²-c²)+(c²-a²)=0 and In the denominator it is (a-b)+(b-c)+(c-a)=0
⇒It will,3(a²-b²)(b²-c²)(c²-a²)/3(a-b)(b-c)(c-a)
So the remainder is
(a+b)(b+c)(c+a) (because we can write (a²-b²)=(a+b)(a-b))
Identity,
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
if x+y+z=0
then,
x³+y³+z³=3xyz
so,
In the question,
(a²-b²)+(b²-c²)+(c²-a²)=0 and In the denominator it is (a-b)+(b-c)+(c-a)=0
⇒It will,3(a²-b²)(b²-c²)(c²-a²)/3(a-b)(b-c)(c-a)
So the remainder is
(a+b)(b+c)(c+a) (because we can write (a²-b²)=(a+b)(a-b))
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