please solve this question ❔❔
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a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-ab-bc-ca]
multiply and divide by 2 in RHS side
=2(a+b+c)/2[a^2+b^2+c^2-ab-bc-ca]
=(a+b+c)/2[2a^2+2b^2+2c^2-2ab-2bc-2ca]
=(a+b+c)/2[(a^2-2ab+b^2)+(b^2-2bc-c^2)+(c^2-2ca+a^2]
=(a+b+c)/2[(a-b)^2+(b-c)^2+(c-a)^2]
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