(a+b)3+(c-a)3-(b+c)3=(b+c)((a+b)2-(a+b)(c-a)+(a-c)2)-(b+c)3=(b+c)(a2+b2+2ab+a2-ac-bc+ab+a2-2ac+c2-b2-c2-2bc)=(b+c)(3a2+3ab-3ac-3bc)=3(b+c)(a-c)(a+b)
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