If $x = 2 + 2^{1 / 3} + 2^{2 / 3}$ then value of $x^{3} - 6 x^{2} + 6 x - 2$ is

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Solution

The correct option is A 0
We have,
$x = 2 + 2^{1 / 3} + 2^{2 / 3}$
$=> x - 2 = 2^{1 / 3} + 2^{2 / 3}$
$=> x - 2 = 2^{1 / 3} (1 + 2^{1 / 3})$
Taking cube on both sides, we get
$=> (x - 2)^{3} = [2^{1 / 3} (1 + 2^{1 / 3})]^{3}$
$=> x^{3} - 8 - 3. x^{2} . 2 + 3. x . 2^{2} = 2 (1 + 2^{1 / 3})^{3}$
$=> x^{3} - 8 - 6 x^{2} + 12 x = 2 (1 + 2 + {3.1}^{2} {.2}^{1 / 3} + {3.1.2}^{2 / 3})$
$=> x^{3} - 6 x^{2} + 12 x - 8 = 2 [3 + {3.2}^{1 / 3} + {3.2}^{2 / 3}]$
$= 6 (x - 1)$ ----(i) $[∵ x = 2 + 2^{\frac{1}{3}} + 2^{\frac{2}{3}} ∴ x - 1 = 1 + 2^{\frac{1}{3}} + 2^{\frac{2}{3}}]$
$= 6 (1 + 2^{1 / 3} + 2^{2 / 3})$
$=> x^{3} - 6 x^{2} + 12 x - 8 = 6 x - 6$
$=> x^{3} - 6 x^{2} + 12 x - 6 x - 8 + 6 = 0$
$=> x^{3} - 6 x^{2} + 6 x - 2 = 0$

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Similar questions

If $x$ = $2 + 2^{\frac{2}{3}} + 2^{\frac{1}{3}}$ , then $x^{3} - 6 x^{2} + 6 x$ =

x = 2 + 2^{1 / 3} + 2^{2 / 3}

x^{3} - 6 x^{2} + 6 x

x = 2 + 2^{\frac{1}{3}} + 2^{\frac{2}{3}}

x^{3} - 6 x^{2} + 6 x =

x = 2 + 2^{2 / 3} + 2^{1 / 3},

x^{3} - 6 x^{2} + 6 x

x = 2 + 2^{\frac{2}{3}}

2^{\frac{1}{3}}

x^{3} - 6 x^{2} + 6 x

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