A) 81
B) 125
C) 343
D) 64
Correct Answer: D
Solution :
In the given numbers, 64 is only number which is a perfect square as well as a perfect cube. Let \[x=\sqrt{6+\sqrt{6+...}}\] \[\Rightarrow \] \[{{x}^{2}}=\sqrt{6+x}\] \[\Rightarrow \] \[{{x}^{2}}-x-6=0\] \[\Rightarrow \] \[{{x}^{2}}-3x+2x-6=0\] \[\Rightarrow \] \[(x-3)(x+2)=0\] \[\Rightarrow \] \[x=3\] \[(\because \,\,x\ne -\,2)\]You need to login to perform this action.
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