frac{1}{2}(a+b+c){(a-b)^{2}+(b-c)^{2}+(c-a)^{ | vedclass

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(B) We know the standard algebraic identity:$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$Multiplying and dividing the right side by $

Vedclass · Accepted Answer

$a^{3}+b^{3}+c^{3}-3abc$

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(B) We know the standard algebraic identity:$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$Multiplying and dividing the right side by $2$,we get:$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)(2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca)$This can be rewritten as:$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)\{(a^{2}-2ab+b^{2})+(b^{2}-2bc+c^{2})+(c^{2}-2ca+a^{2})\}$$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}$Therefore,the given expression is equal to $a^{3}+b^{3}+c^{3}-3abc$.

$\frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}=?$

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