A) 862
B) 863
C) 868
D) 866
Correct Answer: D
Solution :
| \[{{(x+y+z)}^{2}}=144\] |
| \[\therefore \sum{{{x}^{2}}+2\sum{x}y=144}\] |
| \[\Rightarrow \sum{xy=24}\] |
| \[\frac{\sum{xy}}{xyz}=36\] |
| \[\Rightarrow xyz=\frac{2}{3}\] |
| \[\therefore {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=(x+y+z)\,\,(\sum{{{x}^{2}}-\sum{xy)}}\] |
| \[\therefore \sum{{{x}^{3}}-2=12\,\,(96-24)}\] |
| \[\Rightarrow \sum{{{x}^{3}}=866}\] |
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