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It is easy to solve $x^3-1=0$ by having the form $(x-1)(x^2+x+1)=0$
The $3$ roots are: $x=1$, $x=\frac{-1\pm i\sqrt 3}{2}$.

But, solving the equation $x^3+1=0$ by using the similar approach of using the standard approach to solving the quadratic equation & the complex roots does not yield.
$x^3+1=0\implies x^3 -1=-2 \implies (x-1)(x^2+x+1)=-2$ leads to $x-1=-2\implies x=-1$.
The solution $x^2+x+1=-2$ will lead to $x^2+x+3=0$ with roots $x =\frac{-1\pm i \sqrt{11}}{2}$.

But, the answer is $x =-1, \frac{1\pm i \sqrt{3}}{2}$.

jiten
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    Why do you think that $(x - 1)(x^2 + x + 1) = -2$ leads to $x - 1 = -2$? –  Apr 09 '18 at 02:52
  • Substitute $,x=-y,$ then it reduces to the first case. Or just remember that $a^3+b^3=(a+b)(\ldots)$ – dxiv Apr 09 '18 at 02:53
  • @user296602 I feel the error is all four integer combinations need be taken up:(i) $(x-1)=-2 \wedge (x^2+x+1)=1$, (ii) $(x-1)=2 \wedge (x^2+x+1)=-1$, (iii) $(x-1)=1 \wedge (x^2+x+1)=-2$, (iv) $(x-1)=-1 \wedge (x^2+x+1)=2$. I have taken as if both the factors $(x-1=2)\wedge (x^2+x+1=2)$. Also, am looking at all possible solutions, rational, irrational. The given approach applies only for $x3-1=0$ as the both factors are equal to $0$ – jiten Apr 09 '18 at 02:55
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    Above and beyond that, since you're looking at non-integer solutions you also have to consider e.g. $x-1=\sqrt{2}, x^2+x+1=-\sqrt{2}$. Of course this is impossible, but a priori there's no way of knowing that. The fact that $ab=0$ lets you deduce that $a=0$ or $b=0$ is a special property of zero. – Steven Stadnicki Apr 09 '18 at 02:56
  • @jiten I think that you're very confused. You cannot conclude from $ab = -2$ that either $a$ or $b$ are $\pm 1$ or $\pm 2$. I think you've mixed it up with the valid implication that $ab = 0 \implies a = 0$ or $b = 0$. –  Apr 09 '18 at 02:59
  • @StevenStadnicki Please tell that how with out using complex roots can get the given solution, as the book is of 'Imagining Numbers' by Barry Mazur, and is showing the history and am not sure on reading the given part if using anything beyond quadratic equation. – jiten Apr 09 '18 at 02:59
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    Do you realize that $x+1$ is a factor of $x^3+1?$ – saulspatz Apr 09 '18 at 03:01
  • @saulspatz Yes, as $(-1)^3+1=0$. Understood the crux of your comment by the given answer. Thanks. – jiten Apr 09 '18 at 03:04

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Your error is in the line where you say that

... $(x - 1)(x^2 + x + 1) = -2$ leads to $x - 1 = -2$....

This is not true. In general, $ab = -2$ does not imply that $a$ or $b$ are in the set $\{1, 2, -1, -2\}$; for example, we could have $a = \pi$ and $b = -2/\pi$. You seem to have mixed this up with a property of zero. The valid implication is that

$$ab = 0 \implies a = 0 \text{ or } b = 0$$

for complex numbers $a, b$. Using this observation with the factorization

$$x^3 + 1 = (x + 1)(x^2 - x + 1)$$

will lead to a viable solution.