If $A^3=A+I$, then $\det A>0$ 
If $A$ is a $n\times n$ matrix such that $A^3=A+I$, then $\det A>0$.

I don't know how to solve this problem. It's easy to see that $\det A\neq 0$. Suppose $\det A<0$. These are ways that I tried:
$\bullet$ $\det A^3=(\det A)^3<0$, thus $\det(A+I)<0$; $A^2+A+I=A^3+A^2=A^2(A+I)$, thus $\det (A^2+A+I)<0$.
$\bullet$ Let $\lambda_1,\lambda_2,\ldots,\lambda_n$ are all eigenvalues of $A$. Then $\lambda_i^3-\lambda_i$ is eigenvalue of $A^3-A$ for all $i=\overline{1,n}$. But $A^3-A=I$, then $\lambda_i^3-\lambda_i=1$ for all $i$. 
$\bullet$ $\lambda_i^2+\lambda_i+1$ are eigenvalues of $A^2+A+I$. Since $\det (A^2+A+I)<0$, $\displaystyle \prod_{i=1}^n (\lambda_i^2+\lambda_i+1)<0$. 
The problem is I don't know how to deal with complex eigenvalues, so I don't know how to connect all the above things.
Thanks in advance.
 A: Hint
Assuming that the matrix is over $\mathbb{R}$. 
Note that it is obvious that $A$ is invertible, hence $\text{det}A \neq 0$.
The polynomial $p(x)=x^3-x-1$ is an annihilating polynomial for your matrix $A$. Thus the minimal polynomial of $A$ would be either this is or one of its factors. Note that $p(x)$ has only one real root (let us call it) $r$ that lies in the interval $(1,2)$ (note: $r>0$). 
Case 1: If the minimal polynomial for $A$ is $x-r$, then the only real roots of the characteristic polynomial will be $r$ (with multiplicity), the rest will be complex. In which case the determinant of $A$ being the product of eigen values will be positive because the complex roots will occur in conjugate pairs (hence the product will be positive) and the product of $r$ (with multiplicity) will be positive.
Case 2: use a similar argument when minimal polynomial for $A$ is the quadratic factor of $p(x)$.
Case 3: use a similar argument when minimal polynomial for $A$ is $p(x)$.
A: Another solution; assuming that $A$ is with real coefficients, and putting $P(x)=\rm{ det}(A-xI)\in \mathbb{R}[x]$ and $j=\exp(2i\pi/3)$, we compute easily $A^5=A^2+A+I$ hence $${\rm det}(A)^5={\rm det}(A^2+A+I)={\rm det}(A-jI){\rm det}(A-j^2I)=P(j)P(j^2)=|P(j)|^2\geq 0$$ and as $det(A)\not =0$, we are done. 
A: It seems like your hypothesis might be missing something, specifically that $A$ is real. Without that, consider the $1\times 1$ matrix, consisting of one of the complex roots of $p(t)=t^3-t-1$. Its determinant is complex.
Similarly, you can build a diagonal matrix as large as you like, so long as its diagonal entries are all roots of the polynomial, and it will satisfy the equation. If you don't pair all the complex entries up with their conjugate, or else have them in groups of three (their arguments are $\pm 2\pi/3$) you don't have a real valued determinant.
The only way around this, I suspect, is to force that the matrix is real valued, and so the roots appear with their complex conjugate if they are complex, and so their product is positive and real.
eta: Anurag is correct in all particulars, but you do need the assumption that the matrix is real.
A: the eigenvalues are given by $f(\lambda) = \lambda^3 - \lambda - 1 = 0$
claim: the real eigenvalue of $A$ is positive. you can prove the claim by finding the critical numbers $\pm \dfrac{1}{\sqrt 3}$ of $f$ and showing that local max 
$f(-1/\sqrt 3) < 0$ and the only $x$-intercept is positive.
the eigenvalues of $A$ are repetitions of this eigenvalue and complex conjugate pairs. the determinant of $A$ which is the product all the eigenvalues must be positive. 
A: Another way to go at this problem :
Rewrite the equation as $$A(A^2-I)=I$$
Then by definition A is invertiable and $$A^{-1}=A^2-I$$
A: Since $A(A^2 -I)=I,$ taking the determinant on both sides we get $(\det{A}) (\det (A^2 -I)) =1.$ Clearly $\det(A^2 -I) \neq 0$ because if it were then $1$ or $-1$ would be eigenvalues of $A$ and hence $1$ or $ -1$ would be roots of $x^3 -x-1=0,$ a contradiction. Therefore, $\det(A^2 -I) \neq 0$ and $\det A = (\det (A^2-I))^{-1}.$
Now $f(t)= \det(A-tI)$ is a continuous function in $t$ so if $f(-1)$ and $f(1)$ were to have opposite sign by the Intermediate Value Theorem there would exist $c \in (-1, 1)$ such that $f(c) =0.$ Thus $A$ would have an eigenvalue $c \in (-1,1)$ and such a $c$ would satisfy $x^3 -x-1=0.$ This contradicts the fact that $g(x) = x^3 -x-1$ does not change sign on $(-1, 1).$
Thus $f(1)f(-1) >0$ which implies $\det(A^2 -I) >0$ and hence $\det A >0.$
