Prove that
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Using identity: a3 + b3 + c3 - 3abc = (a+b+c)(a2 + b2 + c2 - ab - bc - ca)
(x+y)3 + (y+z)3 + (z+x)3 - 3(x+y) (y+z) (z+x)
= (x+y + y+z + z+x) [(x+y)2 + (y+z)2 + (z+x)2 - (x+y)(y+z) - (y+z)(z+x) - (z+x)(x+y)]
= (2x + 2y + 2z) [x2 + y2 + 2xy + y2 + z2 + 2yz + z2 + x2 + 2zx - xy -xz - y2 -yz -yz - yx -z2 -zx -zx - zy - x2- xy]
= (2x + 2y + 2z) [x2 + y2 + z2 +2xy + 2yz + 2zx - 3xy - 3zx - 3yz - 2y2 - 2z2 - 2x2]
= 2(x+y+z) (x2 + y2 + z2 - xy - yz - zx)
= 2 (x3 + y3 + z3 - 3xyz) (Proved)
hope this is useful for u
(x+y)3 + (y+z)3 + (z+x)3 - 3(x+y) (y+z) (z+x)
= (x+y + y+z + z+x) [(x+y)2 + (y+z)2 + (z+x)2 - (x+y)(y+z) - (y+z)(z+x) - (z+x)(x+y)]
= (2x + 2y + 2z) [x2 + y2 + 2xy + y2 + z2 + 2yz + z2 + x2 + 2zx - xy -xz - y2 -yz -yz - yx -z2 -zx -zx - zy - x2- xy]
= (2x + 2y + 2z) [x2 + y2 + z2 +2xy + 2yz + 2zx - 3xy - 3zx - 3yz - 2y2 - 2z2 - 2x2]
= 2(x+y+z) (x2 + y2 + z2 - xy - yz - zx)
= 2 (x3 + y3 + z3 - 3xyz) (Proved)
hope this is useful for u
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