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prove that (a+b)allcube+(b+c)allcube+(c+a)allcube-3(a+b) (b+c) (c+a)=2(acube+bcube+ccube-3abc)
(a+b)3 + (b+c)3 + (c+a)3 - 3(a+b)(b+c)(c+a)

        ={(a+b)+(b+c)+(c+a)}{((a+b)2+(b+c)2+(c+a)2-(a+b)(b+c)-(c+a)(a+b)}
      
       = 2(a+b+c) (a2+b2+2ab+b2+c2+2bc+c2+a2+2ca-ab-ac-b2-bc-bc-ab-c2-ac-ac-bc-
             
            a2-ab)

       = 2(a+b+c)(a2+b2+c2-ab-bc-ca)

        =2(a3+b3+c3-3abc)


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