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Question Number 217985 by hardmath last updated on 24/Mar/25

x^{3} - 3 x + 1 = 0

The roots of the equation \to a, b, c

Find \to \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = ?

Commented by Frix last updated on 24/Mar/25

suppose a < b < c

\Rightarrow

a = - 2 \cos \frac{π}{9}

b = 2 \sin \frac{π}{18}

c = 2 \cos \frac{2 π}{9}

I {don}^{'} t think

- \sqrt[3]{2 \cos \frac{π}{9}} + \sqrt[3]{2 \sin \frac{π}{18}} + \sqrt[3]{2 \cos \frac{2 π}{9}}

is any “ {nice}^{″} number .

Do you have an exact answer ?

Commented by hardmath last updated on 24/Mar/25

Dear professor, I don't know the answer

Commented by Ghisom last updated on 26/Mar/25

I found a way to solve it \dots

a^{1 / 3} + b^{1 / 3} + c^{1 / 3} = {(3^{5 / 3} - 6)}^{1 / 3}

Answered by ajfour last updated on 26/Mar/25

a b c = - 1

s a y i f {(a b c)}^{1 / 3} = - 1

a + b + c = 0

\Rightarrow (a^{1 / 3} + b^{1 / 3} + c^{1 / 3}) = 0 (m a y b e)

p^{3} + A p - 1 = 0

p^{3} = x

{(x - 1)}^{3} = - A^{3} x

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

\Rightarrow 3 x^{2} - (A^{3} + 6) x + 2 = 0

s a y A^{3} + 6 = t

x = \frac{t}{6} \pm \sqrt{\frac{t^{2}}{36} - \frac{2}{3}}

3 (3 x - 1) - (A^{3} + 6) x^{2} + 2 x = 0

3 (11 x - 3) = (A^{3} + 6) {(A^{3} + 6) x - 2}

x {{(A^{3} + 6)}^{2} - 33} = 2 (A^{3} + 6) - 9

x = \frac{2 t - 9}{t^{2} - 33}

\frac{2 t - 9}{t^{2} - 33} = \frac{t}{6} \pm \sqrt{\frac{t^{2}}{36} - \frac{2}{3}}

\Rightarrow {(\frac{t}{6} - \frac{2 t - 9}{t^{2} - 33})}^{2} = \frac{t^{2} - 24}{36}

\dots

Commented by Frix last updated on 27/Mar/25

a + b + c = 0 \land \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0 is only possible if

c = - {(\sqrt[3]{a} + \sqrt[3]{b})}^{3}

a + b - {(\sqrt[3]{a} + \sqrt[3]{b})}^{3} = 0

- 3 \sqrt[3]{a b} (\sqrt[3]{a} + \sqrt[3]{b}) = 0

\Rightarrow a = 0 \lor b = 0 \lor b = - a [if we stay in R]

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