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# Basics Faster Than Skills

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.

Note by Jason Snow
1 year, 8 months ago

## Comments

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Yes 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 · 1 year, 7 months ago

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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 · 1 year, 7 months ago

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0,same way as rohit. · 1 year, 7 months ago

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Aha, another brilliant person · 1 year, 7 months ago

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0 · 1 year, 7 months ago

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Yeah, did it take you a long time? :D · 1 year, 7 months ago

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Certainly not.... · 1 year, 7 months ago

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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 · 1 year, 7 months ago

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Ha-ha, next time I'll try it, thanks for your nice method :D · 1 year, 7 months ago

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