How can I justify this without determining the determinant? I need to justify the following equation is true:
$$
    \begin{vmatrix}
    a_1+b_1x & a_1x+b_1 & c_1 \\
    a_2+b_2x & a_2x+b_2 & c_2 \\
    a_3+b_3x & a_3x+b_3 & c_3 \\
    \end{vmatrix} = (1-x^2)\cdot\begin{vmatrix}
    a_1 & b_1 & c_1 \\
    a_2 & b_2 & c_2 \\
    a_3 & b_3 & c_3 \\
    \end{vmatrix}
$$
I tried dividing the determinant of the first matrix in the sum of two, so the first would not have $b's$ and the second wouldn't have $a's$. 
Then I'd multiply by $\frac 1x$ in the first column of the second matrix and the first column of the second, so I'd have $x^2$ times the sum of the determinants of the two matrices. 
I could then subtract column 1 to column 2 in both matrices, and we'd have a column of zeros in both, hence the determinant is zero on both and times $x^2$ would still be zero, so I didn't prove anything. What did I do wrong?

Sneaky Solution. . .
The left-hand-side is a polynomial of degree $2$ with zeroes at $x=\pm1$. Hence it has the form $C(1+x)(1-x) = C(1-x^2)$ for some $C \in \mathbb R$. Setting $x=0$ we get  $C=\begin{vmatrix}
    a_1 & b_1 & c_1 \\
    a_2 & b_2 & c_2 \\
    a_3 & b_3 & c_3 \\
    \end{vmatrix}$ as required.
To see $x=\pm1$.  are zeroes  observe for $x=1$ the first and second columns are equal, hence the columns are linearly dependent, and the determinant is zero. For $x=-1$ the first column is the negative of the second, the columns are linearly dependent and the determinant is zero. 
