The roots of the given equation
$27 x^{3} + 21 x + 8 = 0$ are

\frac{1 \pm \sqrt{- 31}}{6}

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- \frac{1}{3}

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- \frac{2}{3}

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\frac{- 1 \pm \sqrt{- 32}}{6}

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Solution

The correct options are
A $\frac{1 \pm \sqrt{- 31}}{6}$
B $- \frac{1}{3}$
Given expression is $27 x^{3} + 21 x + 8 = 0$
$\Rightarrow 27 x^{3} + 1 + 27 x^{2} + 9 x - 1 - 27 x^{2} - 9 x + 21 x + 8 = 0$
$\Rightarrow (27 x^{3} + 1 + 27 x^{2} + 9 x) - 27 x^{2} + 12 x + 7 = 0$
$\Rightarrow {(3 x + 1)}^{3} - (27 x^{2} - 12 x - 7) = 0$
$\Rightarrow {(3 x + 1)}^{3} - (27 x^{2} + 9 x - 21 x - 7) = 0$
$\Rightarrow {(3 x + 1)}^{3} - (9 x (3 x + 1) - 7 (3 x + 1) = 0$
$\Rightarrow {(3 x + 1)}^{3} - (3 x + 1) (9 x - 7) = 0$
$\Rightarrow (3 x + 1) ({(3 x + 1)}^{2} - (9 x - 7)) = 0$
$\Rightarrow (3 x + 1) (9 x^{2} + 1 + 6 x - 9 x + 7) = 0$
$\Rightarrow (3 x + 1) (9 x^{2} - 3 x + 8) = 0$
$\Rightarrow (3 x + 1) = 0 \Rightarrow x = - \frac{1}{3}$
Also $9 x^{2} - 3 x + 8 = 0$
Using quadratic formula, we have
$x = \frac{3 \pm \sqrt{9 - 4 (9) (8) 2}}{2 (9)} = \frac{3 \pm \sqrt{- 567}}{18} x = \frac{3 \pm 3 \sqrt{- 63}}{6} = \frac{1 \pm \sqrt{- 63}}{6}$
So, the values of $x$ are $\frac{1 + \sqrt{- 63}}{6}, \frac{1 - \sqrt{- 63}}{6}$ and $- \frac{1}{3}$ .

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