Question : If $x=2-2^{\frac{1}{3}}+2^{\frac{2}{3}}$, then find the value of $x^3-6 x^2+18 x$.
Option 1: 40
Option 2: 33
Option 3: 45
Option 4: 22
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Correct Answer: 22
Solution :
$x=2-2^{\frac{1}{3}}+2^{\frac{2}{3}}$
⇒ $(x-2) = (2^{\frac{2}{3}}-2^{\frac{1}{3}})$
Cubing both sides,
⇒ $(x-2)^3 = (2^{\frac{2}{3}}-2^{\frac{1}{3}})^3$
⇒ $x^3 - 6x^2 +12x-8 = 2^2 -3 × 2^{\frac{2}{3}} × 2^{\frac{1}{3}}(2^{\frac{2}{3}}-2^{\frac{1}{3}}) - 2$
⇒ $x^3 - 6x^2 +12x-8 = 4 - 3 × 2(x-2) - 2$
⇒ $x^3 - 6x^2 + 12x = 10 - 6x +12$
⇒ $x^3 - 6x^2 + 12x + 6x = 22$
Hence, the correct answer is 22.
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