How can I justify this without determining the determinant?

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 $$ \det(C) = \begin{vmatrix} 1 & x & 0 \\ x & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} = \begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = 1 \cdot 1 - x \cdot x = 1-x^2. $$

\begin{align} &\phantom {=}\,\ \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} \\ &= \begin{vmatrix} a_1 & a_1x+b_1 & c_1 \\ a_2 & a_2x+b_2 & c_2 \\ a_3 & a_3x+b_3 & c_3 \end{vmatrix} + \begin{vmatrix} b_1x & a_1x+b_1 & c_1 \\ b_2x & a_2x+b_2 & c_2 \\ b_3x & a_3x+b_3 & c_3 \end{vmatrix} \\&= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + x \begin{vmatrix} b_1 & a_1x & c_1 \\ b_2 & a_2x & c_2 \\ b_3 & a_3x & c_3 \end{vmatrix} \\&= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + x^2 \begin{vmatrix} b_1 & a_1 & c_1 \\ b_2 & a_2 & c_2 \\ b_3 & a_3 & c_3 \end{vmatrix} \\&= 1\cdot \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + (-1) x^2 \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & 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}. \end{align}

The determinant is a polynomial of order 2 in $x$, $D(x)$, where the coefficients depend of the $a_i$, $b_i$ and $c_i$.

We know its two roots $1$ and $-1$, as the determinant is obviously null in these cases: two identical columns or one column the inverse of another one.

Therefore $$ D(x) = \lambda (1-x^2)$$

Where $\lambda$ depends of the $a_i$, $b_i$ and $c_i$.

Finally, the multiplicative term is given by $x=0$ :

$$D(0) =\lambda = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}$$