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Definition of Functions

If x = 2 + ...

Question

If $x = 2 + 2^{\frac{2}{3}}$ + $2^{\frac{1}{3}}$ then the value of $x^{3} - 6 x^{2} + 6 x$ is :

A

3

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B

2

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C

1

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D

none

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Solution

The correct option is D $2$
We have, $x = 2 + 2^{\frac{2}{3}} + 2^{\frac{1}{3}}$

$\Rightarrow x - 2 = 2^{\frac{2}{3}} + 2^{\frac{1}{3}}$

$\Rightarrow {(x - 2)}^{3} = {(2^{\frac{2}{3}} + 2^{\frac{1}{3}})}^{3}$

By using, $(a - b)^{3} = a^{3} - b^{3} - 3 a b (a - b)$ and $(a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b)$

Now
$x^{3} - 8 - 6 x^{2} + 12 x = 4 + 2 + 6 (x - 2)$
$x^{3} - 8 - 6 x^{2} + 12 x = 6 + 6 x - 12$
$x^{3} - 6 x^{2} + 6 x = 2$

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Similar questions

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

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

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

then the value of

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

, then the value of

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

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

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