Since you're using a $3\times3$ matrix you can use this system:

It follows directly from the definition of the determinant which is quite the hairy function so bear with me.
There are generally 2 ways to introduce determinants, one is by defining the formula and stating the properties, the other is by stating the properties and deriving the formula. I prefer the latter, but entire chapters have been dedicated to this, so we'll be skipping most steps and I'll just introduce the formula. Hoffman and Kunze's Linear Algebra has a great chapter on determinants.
First, let's call $\sigma$ a permutation, and we'll define $sgn(\sigma)=(-1)^z$ where $z$ stands for the amount of "switches" that have to happen in order to arrive at a permutation. Example:
If we consider a series $(1,2,3)$, then if we change this to $(1,3,2), \quad sgn(\sigma)=-1$. Change it to $(3,1,2)$ and $sgn(\sigma)=1$.
Now the determinant is defined as:
$$det(A)=\sum_\sigma(sgn(\sigma))A_{(1,\sigma1)},...,A_{(n,\sigma n)}$$
That is to say the sum of all permutations where the horizontal positions of the terms within our product get determined by the specific permutation.
Let us consider a $3\times3$ matrix. Then all possible permutations are:
$$(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)$$
(It is no coincidence that this coincides with $3!$)
Can you use this definition to show that $$det \begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}=0$$
?
This method, though clunky, will really deepen your understanding of what a determinant is and how it works. Cofactor expansion will generally be easier when it comes to a quick calculation, but the reason it works can be found in this formula.