If a=25, b=15, c=-10; then the value of frac{ | vedclass

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(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.This can also be written as: $a^{3}+b^{3}+c^{3}-3ab

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$15$

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(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.This can also be written as: $a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$.Substituting this into the given expression:$\frac{\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}} = \frac{1}{2}(a+b+c)$.Given $a=25, b=15, c=-10$,we have:$\frac{1}{2}(25+15-10) = \frac{1}{2}(30) = 15$.

If $a=25, b=15, c=-10$; then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$ is

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