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Higher Order Derivatives

If x=1+21/3...

Question

If $x = 1 + 2^{1 / 3} + 2^{2 / 3}$ then find the value of $x^{3} - 3 x^{2} - 3 x - 1$ .

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Solution

$x = 1 + 2^{1 / 3} + 2^{2 / 3}$
$x - 1 = (2^{1 / 3} + 2^{2 / 3})$
$(x - 1)^{3} = (2^{1 / 3} + 2^{2 / 3})^{3}$
$x^{3} - 3 x^{2} + 3 x - 1 = (2^{1 / 3})^{3} + (2^{2 / 3})^{3} + {3.2}^{1 / 3} {.2}^{2 / 3} (2^{1 / 3} + 2^{2 / 3})$
$x^{3} - 3 x^{2} + 3 x - 1 = 2 + 2^{2} + {3.2}^{1} (x - 1)$
$x^{3} - 3 x^{2} + 3 x - 1 = 6 + 6 (x - 1)$
$x^{3} - 3 x^{2} + 3 x - 1 = 6 x$
$x^{3} - 3 x^{2} - 3 x - 1 = 0$

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

then the value of

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

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

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then the value of

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

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

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then find the value of

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