How many distinct real roots are possible for the below equation?
$x^{6} - 26 x^{3} - 27 = 0$

-1, 27

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1, -27

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-1, 3

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1, -3

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Solution

The correct option is B 1, -27
Let $x^{3} = y$

$y^{2} - 26 y - 27 = 0$

$y^{2} - 27 y + y - 27 = 0$

$y (y - 27) + 1 (y - 27) = 0$

$(y + 1) (y - 27) = 0$

The roots are y = -1, y = 27

$x^{3} = - 1, x = - 1$

$x^{3} = 27, x = 3$

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x^{6} - 26 x^{3} - 27

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