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Laws of Exponents

If x=2+21/3...

Question

If $x = 2 + 2^{1 / 3} + 2^{2 / 3}$ , then $x^{3} - 6 x^{2} + 6 x =$ .............

A

2

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B

1

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C

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D

3

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Solution

The correct option is A $2$
$x = 2 + 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$
$\Rightarrow (x - 2)^{3} = (2^{\frac{1}{3}} + 2^{\frac{2}{3}})^{3}$
$\Rightarrow x^{3} - 6 x^{2} + 12 x - 8 = 2 + 4 + 3 \times 2 [2^{\frac{1}{3}} + 2^{\frac{2}{3}}]$
$x^{3} - 6 x^{2} + 6 x + 6 x = 14 + 6 [2^{\frac{1}{3}} + 2^{\frac{2}{3}}]$
$\Rightarrow x^{3} - 6 x^{2} + 6 x = 14 + 6 [2^{\frac{1}{3}} + 2^{\frac{2}{3}}] - 12 - 6 [2^{\frac{1}{3}} + 2^{\frac{2}{3}}]$
$\Rightarrow x^{3} - 6 x^{2} + 6 x = 2$

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Similar questions

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

2^{\frac{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 - 2

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

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

then prove that

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

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