A) 32
B) 14
C) 41
D) 23
Correct Answer: D
Solution :
Let the number be \[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca}=\_\_\_\_\_\_\_\_\] Sum of digits \[a+b+c\] ?.(1) Further, \[10x+y+9=10y+x\] \[\Rightarrow \] \[9x-9y=-9\] \[\Rightarrow \] \[12x+11y=57,\] ?.(2) Adding (1) and (2), we have \[{{\left( \frac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b}}\times {{\left( \frac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c}}\times {{\left( \frac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a}}\] \[a=\frac{1}{2-\sqrt{3}},b=\frac{1}{2+\sqrt{3},}\] \[{{\left( \frac{a+b}{a-b} \right)}^{2}}\] From (1), \[y=5-2=3\] Hence, the number\[=10(2)+3=23\]
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