A) 0
B) 1
C) 2
D) 6
Correct Answer: A
Solution :
[a] \[x=2+{{2}^{1/3}}+{{2}^{2/3}}\] \[x-2={{2}^{1/3}}+{{2}^{2/3}}={{2}^{1/3}}(1+{{2}^{1/3}})\] \[\Rightarrow \] \[{{(x-2)}^{3}}={{[{{2}^{1/3}}(1+{{2}^{1/3}})]}^{3}}\] \[\Rightarrow \] \[{{x}^{3}}-8-3.{{x}^{2}}.2+3.x{{.2}^{2}}=2{{(1+{{2}^{1/3}})}^{3}}\] \[\Rightarrow \]\[{{x}^{3}}-8-6{{x}^{2}}+12x=2\,\,(1+2+{{3.1}^{2}}{{.2}^{1/3}}+{{3.1.2}^{2/3}})\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+12x-8=2\,\,[3+{{3.2}^{1/3}}+{{3.2}^{2/3}}]\] \[=\,\,\,6(1+{{2}^{1/3}}+{{2}^{2/3}})\] \[=6(x-1)\] ? (i) \[\left[ \begin{matrix} \because & x=2+{{2}^{\frac{1}{3}}}+{{2}^{\frac{2}{3}}} \\ \therefore & x-1=1+{{2}^{\frac{1}{3}}}+{{2}^{\frac{2}{3}}} \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+12x-8=6x-6\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+12x-6x-8+6=0\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+6x-2=0\]
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