If $x^{3} - {(x + 1)}^{2} = 2001$ then the value of x is

14

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13

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10

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None of these

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Solution

The correct option is B $13$
given, $x^{3} - (x + 1)^{2} = 2001$

$x^{3} - x^{2} - 2 x - 1 = 2001$

$x^{3} - x^{2} - 2 x - 2002 = 0$

$x^{3} - 2197 - x^{2} + 169 - 2 x + 26 = 0$

$(x^{3} - 13^{3}) - (x^{2} - 13^{2}) - 2 (x - 13) = 0$

$(x - 13) (x^{2} + 13 x + 169) - (x - 13) (x + 13) - 2 (x - 13) = 0$

$(x - 13) [x^{2} + 13 x + 169 - x - 13 - 2] = 0$

$(x - 13) [x^{2} + 12 x + 154] = 0$

$x = 13$ or $x^{2} + 12 x + 154 = 0$

If $x^{2} + 12 x + 154 = 0$

$b^{2} - 4 a c = - 472 < 0$

Hence, $x = 13$

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