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If x = $$2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}},$$ then find the value of x3 - 6x2 + 18x.
Answer & Solution
Correct Answer:
Option
D
$$\eqalign{
& x = 2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}} \cr
& x - 2 = {2^{\frac{2}{3}}} - {2^{\frac{1}{3}}} \cr
& {\left( {x - 2} \right)^3} = {\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right)^3} \cr
& {x^3} - 8 - 3 \times 2x\left( {x - 2} \right) = 4 - 2 - 3 \times {2^{\frac{2}{3}}} \times {2^{\frac{1}{3}}}\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right) \cr
& {x^3} - 8 - 6{x^2} + 12x = 4 - 2 - 6\left( {x - 2} \right) \cr
& {x^3} - 8 - 6{x^2} + 12x = 2 - 6x + 12 \cr
& {x^3} - 6{x^2} + 12x + 6x = 2 + 12 + 8 \cr
& {x^3} - 6{x^2} + 18x = 22 \cr} $$
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