A) \[30\]
B) \[-15\]
C) \[-\,30\]
D) \[15\]
Correct Answer: D
Solution :
| [d] We know that\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] \[=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] \[=\frac{1}{4}(a+b+c)\{{{(a-b)}^{2}}+{{c}^{2}}-2ab\] \[-2bc-2ca\}\] Now, \[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{(a+b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}}\] \[(a+b+c)\{{{(a-b)}^{2}}+{{(b-c)}^{2}}\] \[=\frac{1}{2}\frac{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{(c-a)}^{2}}\}}{\{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}\}}\] \[=\frac{a+b+c}{2}=\frac{25+15-10}{2}=\frac{30}{2}=15\] |
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