Solve:
$x^{3} - 4 x^{2} - x + 1 = (x - 2)^{3}$

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Solution

$x^{3} - 4 x^{2} - x + 1 = (x - 2)^{3}$
using identify $(a - b)^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}$
we get
$x^{3} - 4 x^{2} - x + 1 = x^{3} - 8 - 6 x^{2} + 12 x$
$6 x^{2} - 4 x^{2} - x - 12 x + 1 + 8 = 0$
$2 x^{2} - 13 x + 9 = 0$
$x = \frac{+ 13 \pm \sqrt{169 - 72}}{4} = \frac{13 \pm \sqrt{97}}{4}$
Hence $x = \frac{13 + \sqrt{97}}{4}$ & $\frac{13 - \sqrt{97}}{4}$

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