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Prove that : a^3+b^3+c^3-3a b c=1/2(a+b+...

Prove that : `a^3+b^3+c^3-3a b c=1/2(a+b+c)"{"a-b")"^2+(b-c)^2+(c-a)^2}`

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we need to prove that the left hand side (L.H.S) of the equation is equal to right hand side (R.H.S).
Therefore R.H.S,
`=1/2(a+b+c){"a-b")"^2+(b-c)^2+(c-a)^2}`
`=1/2(a+b+c){a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac}`
`=1/2(a+b+c)2{a^2+b^2+c^2-ab-bc-ca}`
`=(a+b+c){a^2+b^2+c^2-ab-bc-ca}`
`=a^3+ab^2+ac^2-a^2b-abc-ca^2+b^3+ba^2+bc^2-b^2a-abc-cb^2+ca^2+cb^2+c^3-abc-bc^2-c^2a`
after simplification,
`=a^3+b^3+c^3-3abc`
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