For $x^3+px+q=0$, one of the solutions is $\sqrt 3-1$. Find the value of rational number $p$ and $q$, and other two solutions of the equation. 
I cannot find any error from my answer. What’s wrong?

If $\sqrt{3} -1$ is a root of $x^3+px+q$, it doesn't necessarily follow that $\sqrt{3}+1$ is also a root.
In fact, if $\sqrt{3} -1$ is a root then $\sqrt{3}+1$ cannot be a root, assuming $p,q \in \mathbb{Q}$.
Indeed
$$\alpha = \sqrt{3} + 1 \implies \alpha-1 = \sqrt{3} \implies (\alpha-1)^2 = 3 \implies \alpha^2-2\alpha-2 = 0$$
so $x^2-2x-2$ is the minimal polynomial of $\sqrt{3} + 1$ over $\mathbb{Q}$.
On the other hand
$$\beta = \sqrt{3} - 1 \implies \beta+1 = \sqrt{3} \implies (\beta+1)^2 = 3 \implies \beta^2+2\beta -2 = 0$$
so $x^2+2x-2$ is the minimal polynomial of $\sqrt{3} - 1$ over $\mathbb{Q}$.
If a polynomial $f$ with rational coefficients satisfies $f(\sqrt{3} + 1) = f(\sqrt{3} - 1)  = 0$ then $x^2-2x-2$ and $x^2+2x-2$ both divide $f$, and they are irreducible so $\deg f \ge 4$.

However, what you can deduce is that $-\sqrt{3} - 1$ is necessary also a root of $x^3+px+q$. This is because the minimal polynomial of $\sqrt{3}-1$,  which is $x^2+2x-2$, must divide $x^3+px+q$. And $-\sqrt{3} - 1$ happens to be a root of $x^2+2x-2$.
Therefore
$$(\sqrt{3}-1)^3 + p(\sqrt{3}-1) + q = 0$$
$$(-\sqrt{3}-1)^3 + p(-\sqrt{3}-1) + q = 0$$
which yields
$$-10+6\sqrt{3} + p(\sqrt{3}-1) + q  = 0$$
$$-10+6\sqrt{3} + p(\sqrt{3}-1) + q  = 0$$
Subtracting them gives $p = -6$ and adding them gives $q = 4$.
Therefore, your polynomial is $x^3 - 6x + 4 = (x-2)(x^2+2x-2)$ so the third root is $2$.
All of this is unnecessary, of course. The solution from the book is the shortest.
