Factorize \((b-c)^{6}+(c-a)^{6}+(a-b)^{6}-9(a-b)^{2}(b-c)^{2}(c-a)^{2}-2(a-b)^{3}(a-c)^{3}-2(b-c)^{3}(b-a)^{3}-2(c-a)^{3}(c-b)^{3}\)

- When you got the answer you must be surprised, feeling cheated.
- Think about the title carefully, in that there are two ways to solve it.

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## Comments

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TopNewestYes indeed...it looks quite symmetric doesn't it?

Let \(a-b=x, b-c=y, c-a=z\)

Then the expression becomes:

\(x^6+y^6+z^6-9x^2y^2z^2+2x^3y^3+2y^3z^3+2z^3x^3\)

\(= (x^3+y^3+z^3)^2-(3xyz)^2\)

\(= (x^3+y^3+z^3-3xyz)(x^3+y^3+z^3+3xyz)\)

Now we note that \(x+y+z=0\) which means that \(x^3+y^3+z^3=3xyz\) and our whole expression becomes 0

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That's why I said you'll be shocked :D, your method is faster than those I thought because I, in fact, want you guys to get it simplified step by step first, obviously you didn't jump into my trap :P

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0,same way as rohit.

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Aha, another brilliant person

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0

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Yeah, did it take you a long time? :D

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Certainly not....

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My advice to all regarding such long questions is to first try your luck with 0,and if it doesn't work,move to the next one

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