Solving $\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$ for real $x$ 
Solve the equation in the Real number system:
$$\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$$

I have attempted using $(A-B)^3 = A^3 - B^3 - 3.A.B.(A-B)$ with $A = \sqrt[3]{x+1}$ , $B = \sqrt[3]{x-1}$ and $(A-B) = \sqrt[3]{x^2-1}$, however I end up getting $-3.(\sqrt[3]{x^2-1})^2 = x^2-3$ which is not equivalent to the original one since 0 is solution only of $-3.(\sqrt[3]{x^2-1})^2 = x^2-3$. Could someone explain to me why does it not work here?
By the way, the answer according to the book is $\left\lbrace \frac{\sqrt 5}{2},-\frac{\sqrt 5}{2} \right\rbrace$ but I could not get there.
 A: We can use $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
$a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only.
Since  $$\sqrt[3]{x+1}=- \sqrt[3]{x-1} =-\sqrt[3]{x^2-1}$$ gives $x=0$ and $0$ is not a root of the equation, we need to remove the number $0$ and we obtain:
$$x+1-(x-1)-(x^2-1)-3\sqrt[3]{x+1}\cdot\sqrt[3]{x-1}\cdot\sqrt[3]{x^2-1}=0$$ or
$$3-x^2=3\sqrt[3]{(x^2-1)^2}$$ or $$x^2(x^4+18x^2-27)=0,$$ which gives $$\left\{\sqrt{6\sqrt3-9},-\sqrt{6\sqrt3-9}\right\}$$
A: Put $a=\sqrt[3]{x+1}$ and $b= - \sqrt[3]{x-1}$.
Then $a+b=-ab$, hence $a$ satisfy $a^2+pa+p=0$ where $p=ab$ from which
$$a=\frac{-p\pm\sqrt{p^2-4p}}2$$
Consequently,
\begin{align}
x
&=a^3-1\\
&=-pa(a+1)-1\\
&=-p(-pa-p+a)-1\\
&=(p-1)\frac{-(p^2-2p-2)\pm p\sqrt{p^2-4p}}2
\end{align}
On the other hand, we have $a^3+b^3=2$.
Cubing $a+b=-ab$ gives $a^3+b^3+3ab(a+b)=-a^3b^3$ from which $2-3a^2b^2=-a^3b^3$.
If $p=ab$, then
$$0=p^3-3p^2+2=(p-1)(p^2-2p-2)$$
from which $p=1$ and $p=1\pm\sqrt 3$.
Excluding $p=1$ (which gives $x=0$) we get
$$x=\pm\frac{(p+2)\sqrt 2\sqrt{1-p}}2$$
hence $p=1-\sqrt 3$ and
$$x=\pm\frac{(3-\sqrt 3)\sqrt 2\sqrt[4]3}2$$
