Verify that x^3+y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]
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1
Answer:
We have, L.H.S.=x
3
+y
3
+z
3
−3xyz
=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx) [by poynomial identity]
=
2
1
(x+y+z)(2x
2
+2y
2
+2z
2
−2xy−2yz−2zx)
=
2
1
[(x+y+z)(x−y)
2
+(y−z)
2
+(z−x)
2
]
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