Prove that -: a + b + c? –3abc = < (a + b + c) *{(a−b)2 + (6 –c)2 + (c-a)?}
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We know,
a
3
+b
3
+c
3
−3abc=(a+b+c)(a
2
+b
2
+c
2
−ab−bc−ca)
Putting a+b+c=0 on RHS, we get,
a
3
+b
3
+c
3
−3abc=0
⟹a
3
+b
3
+c
3
=3abc, Hence proved.
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