A) \[1\frac{3}{8}\]
B) \[3\frac{3}{8}\]
C) \[2\frac{3}{8}\]
D) 0
Correct Answer: D
Solution :
[d] \[{{(2a-\,1)}^{2}}+{{(4b-\,3)}^{2}}+{{(4c+5)}^{2}}=0\] \[\begin{matrix} =2a-1=0; & 4b-3=0; & 4c+5=0 \\ a=\frac{1}{2} & b=\frac{3}{4}; & c=\frac{-5}{4} \\ \end{matrix}\] \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-\,3\,\,abc\]\[=(a+b+c)\,\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-\,ab-\,bc-\,ca)\]\[But\,\,a+b+c=0\] \[So,\,\,{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-\,3\,\,abc=0\] \[So,\,\,\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-\,3\,\,abc}{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}=0\]
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