A) 22
B) 33
C) 40
D) 45
Correct Answer: C
Solution :
\[x=2-{{2}^{\frac{1}{3}}}+{{2}^{\frac{2}{3}}}\]\[\Rightarrow \]\[x-2={{2}^{\frac{2}{3}}}-{{2}^{\frac{1}{3}}}\] On cubing \[{{x}^{3}}-3{{x}^{2}}\times 2+3x\times 4-8={{\left( {{2}^{\frac{2}{3}}} \right)}^{3}}-{{\left( {{2}^{\frac{1}{3}}} \right)}^{3}}\]\[-3\cdot {{2}^{\frac{2}{3}}}\cdot {{2}^{\frac{1}{3}}}\left( {{2}^{\frac{2}{3}}}-{{2}^{\frac{1}{3}}} \right)\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+12x-8=4-2-6(x-2)\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+12x-8=2-6x+12\] \[={{x}^{3}}-6{{x}^{2}}+18x+18=2+12+8+18=40\]
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