Assume that $A$ is invertible, and prove that the unique solution to the solution $Ax = b$ is given by
$$x_j = \frac{\det(B[j])}{\det(A)} $$
where A the matrix whose columns are denoted by the column vectors $c_1,...,c_n$ and where $B[j]$ is obtained by replacing the jth column of A by the vector $b$, viewed as the column vector.
I am really not sure how to prove this statement.
Write $A = (c_1,\dotsc,c_n)$ and let $x = (x_1,\dotsc,x_n)^\top$ be a solution, i. e. $Ax = b$. Then $b = \sum_{i=1}^nx_i\cdot c_i$. Using linearity of $\det$ in the $j$-th column and that the determinant vanishes, whenever two columns coincide, we compute \begin{align*} \det B[j] &=\det\left(c_1,\dotsc,c_{j-1}, \sum_{i=1}^nx_i c_i, c_{j+1}, \dotsc,c_n\right)\\ &= \sum_{i=1}^n x_i\det(c_1,\dotsc,c_{j-1},c_i,c_{j+1},\dotsc,c_n)\\ &= x_j\cdot \det(A). \end{align*} Hence $x_j = \frac{\det B[j]}{\det A}$ for all $j=1,\dotsc,n$.
Hint:
We have $AX = B \iff X = A^{-1}B = \frac{adj(A)}{det(A)}B$
Now, look at the element $x_i =(X)_{i1}$ using the definition and properties of matrix multiplication and adjugate matrix.