A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
\[x={{2}^{\frac{1}{3}}}+{{2}^{\frac{1}{3}}}\] On cubing both sides. \[{{x}^{3}}={{\left( {{2}^{\frac{1}{3}}}+{{2}^{-\frac{1}{3}}} \right)}^{3}}\] \[\Rightarrow \] \[{{x}^{3}}={{\left( {{2}^{\frac{1}{3}}} \right)}^{3}}+{{\left( {{2}^{-\frac{1}{3}}} \right)}^{3}}+3\times {{2}^{\frac{1}{3}}}\times {{2}^{-\frac{1}{3}}}\left( {{2}^{\frac{1}{3}}}+{{2}^{\frac{1}{3}}} \right)\] \[\Rightarrow \] \[{{x}^{3}}=2+{{2}^{-1}}+3x\] \[[{{a}^{m}}\times {{a}^{-m}}={{a}^{0}}=1]\] \[\Rightarrow \] \[{{x}^{3}}=2+\frac{1}{2}+3x\] \[\Rightarrow \] \[2{{x}^{3}}=4+1+6x\] \[\Rightarrow \] \[2{{x}^{3}}-6x-5=0\]
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