A) 48
B) 52
C) 64
D) None of these
Correct Answer: B
Solution :
\[\therefore \] \[{{m}^{2}}-4m+1=0\] \[\Rightarrow \] \[{{m}^{2}}+1=4m\] \[\Rightarrow \] \[\frac{{{m}^{2}}+1}{m}=4\] \[\Rightarrow \] \[m+\frac{1}{m}=4\] ?(i) \[\therefore \] \[{{\left( m+\frac{1}{m} \right)}^{3}}={{m}^{3}}+\frac{1}{{{m}^{3}}}+3\,\left( m+\frac{1}{m} \right)=64\] [from Eq. (i)] \[\Rightarrow \] \[{{m}^{3}}+\frac{1}{{{m}^{3}}}=64-12=52\]
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