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If a+b+c=12 and A2+b2+C2=90 find the value of A3+b3+c3-3abc

If a+b+c=12 and A2+b2+C2=90 find the value of A3+b3+c3-3abc

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Prove that a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)

Prove that a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)

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Prove that a3+b3+c3-3abc= (a+b+c)(a2+b2+c2-ab-bc-ca)=1/2(a+b+c)(a-b)2+(b-c)2+(c-a)2

Prove that a3+b3+c3-3abc= (a+b+c)(a2+b2+c2-ab-bc-ca)=1/2(a+b+c)(a-b)2+(b-c)2+(c-a)2

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proof a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ac)

proof a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ac)

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Special Identities: a3 + b3 + c3 - 3abc

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