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?

For another solution, note that
$$
\underbrace{\begin{bmatrix}
    a_1+b_1x & a_1x+b_1 & c_1 \\
    a_2+b_2x & a_2x+b_2 & c_2 \\
    a_3+b_3x & a_3x+b_3 & c_3 \\
\end{bmatrix}}_{A}
=
\underbrace{\begin{bmatrix}
    a_1 & b_1 & c_1 \\
    a_2 & b_2 & c_2 \\
    a_3 & b_3 & c_3 \\
\end{bmatrix}}_{B}
\underbrace{\begin{bmatrix}
    1 & x & 0 \\
    x & 1 & 0 \\
    0 & 0 & 1 \\
\end{bmatrix}}_{C}
$$
and therefore $\det(A) = \det(BC) = \det(B)\det(C)$. From there, it's enough to check that 
$$
   \det(C) = \begin{vmatrix}
    1 & x & 0 \\
    x & 1 & 0 \\
    0 & 0 & 1 \\
\end{vmatrix} = \begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = 1 \cdot 1 - x \cdot x = 1-x^2.
$$
