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