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?
 A: 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 
$$
   \det(C) = \begin{vmatrix}
    1 & x & 0 \\
    x & 1 & 0 \\
    0 & 0 & 1 \\
\end{vmatrix} = \begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = 1 \cdot 1 - x \cdot x = 1-x^2.
$$
A: 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. 
A: \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}
A: 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 $x=0$ : 
$$D(0) =\lambda = \begin{vmatrix}
    a_1 & b_1 & c_1 \\
    a_2 & b_2 & c_2 \\
    a_3 & b_3 & c_3 \\
    \end{vmatrix}$$
A: An alternative way of looking at the accepted answer, and justifying the steps more (to answer the questions its comments), is considering the columns as vectors (which I now notice @Semiclassical did in their comment), $$\mathbf{a} = \begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}$$
and same for $\mathbf{b}$ and $\mathbf{c}$. Now we should know determinants are "multilinear", so for any further vectors $\mathbf{v}, \mathbf{w}, \mathbf{z}$ and any scalar $x$, $$\mathrm{det}\bigl( (\mathbf{v}+\mathbf{w}), \mathbf{z}, \mathbf{c}\bigr) = \mathrm{det}\bigl( \mathbf{v}, \mathbf{z}, \mathbf{c}\bigr) +\mathrm{det}\bigl( \mathbf{w}, \mathbf{z}, \mathbf{c}\bigr) \\ 
\mathrm{det}\bigl(\mathbf{v}, (\mathbf{w}+\mathbf{z}), \mathbf{c}\bigr) = \mathrm{det}\bigl( \mathbf{v}, \mathbf{w}, \mathbf{c}\bigr) +\mathrm{det}\bigl( \mathbf{v}, \mathbf{z}, \mathbf{c}\bigr) \\ 
\mathrm{det}\bigl( x\,\mathbf{v}, \mathbf{z}, \mathbf{c}\bigr) = x.\mathrm{det}\bigl( \mathbf{v}, \mathbf{z}, \mathbf{c}\bigr)\\ 
\mathrm{det}\bigl( \mathbf{v}, x\,\mathbf{z}, \mathbf{c}\bigr) = x.\mathrm{det}\bigl( \mathbf{v}, \mathbf{z}, \mathbf{c}\bigr) $$
Knowing this, we know (e.g., from $\mathbf{v}=\mathbf{w}+(\mathbf{v}-\mathbf{w})$ on above) also $$\mathrm{det} \bigl(  \mathbf{v}, \mathbf{w}, \mathbf{c} \bigr) = -\mathrm{det} \bigl(  \mathbf{w}, \mathbf{v}, \mathbf{c} \bigr)\\
\mathrm{det} \bigl(  \mathbf{v}, \mathbf{v}, \mathbf{c} \bigr) = 0$$
This suffices to straightforwardly work out the equality (I'm doing the first matrix column on the first line, then the second column of both on second line): 
$$\require{cancel}\mathrm{det}\bigl( (\mathbf{a}+x\,\mathbf{b}), (x\,\mathbf{a}+\mathbf{b}), \mathbf{c}\bigr) \\ 
= \mathrm{det}\bigl( \mathbf{a}, (x\,\mathbf{a}+\mathbf{b}), \mathbf{c}\bigr) + x.\mathrm{det}\bigl( \mathbf{b}, (x\,\mathbf{a}+\mathbf{b}), \mathbf{c}\bigr)\\
= x.\cancel{\mathrm{det}\bigl( \mathbf{a}, \mathbf{a}, \mathbf{c}\bigr)} +\mathrm{det}\bigl( \mathbf{a}, \mathbf{b}, \mathbf{c}\bigr) + x.\Bigl(x.\mathrm{det}\bigl( \mathbf{b}, \mathbf{a}, \mathbf{c}\bigr)+\cancel{\mathrm{det}\bigl( \mathbf{b}, \mathbf{b}, \mathbf{c}\bigr)}\Bigr)\\
= \mathrm{det}\bigl( \mathbf{a}, \mathbf{b}, \mathbf{c}\bigr) + x^2.\mathrm{det}\bigl( \mathbf{b}, \mathbf{a}, \mathbf{c}\bigr)\\
= \mathrm{det}\bigl( \mathbf{a}, \mathbf{b}, \mathbf{c}\bigr) - x^2.\mathrm{det}\bigl( \mathbf{a}, \mathbf{b}, \mathbf{c}\bigr) = (1- x^2).\mathrm{det}\bigl( \mathbf{a}, \mathbf{b}, \mathbf{c}\bigr)\\$$
The advantage I feel of this notation/approach is that you can see it extends to higher dimensions (adding columns $\mathbf{d}$, $\mathbf{e}$, ... ) without onerous notation. Also, you could replace in the above each "det(...)" by "$f$(...)", so it holds for any multilinear function $f$. 
Compared to the highest-scoring answer: That answer is more elegant, but was found by working backwards from the solution, I feel (but it works equally well in each dimension), and needs a bit more skill in matrices.
A: Let the expression be $d(x)$. $d(x)$ is clearly a quadratic polynomial in $x$.
We observe
$$d(0)=\Delta,\\d(-1)=d(1)=0$$
so that
$$d(x)=(1-x^2)\Delta.$$
(This is the Lagrangian polynomial by the three points.)
A: You can do the column operation: $C_1-xC_2\to C_1$:
$$\begin{vmatrix}
    a_1+b_1x & a_1x+b_1 & c_1 \\
    a_2+b_2x & a_2x+b_2 & c_2 \\
    a_3+b_3x & a_3x+b_3 & c_3 \\
    \end{vmatrix} = \begin{vmatrix}
    (1-x^2)a_1 & a_1x+b_1 & c_1 \\
    (1-x^2)a_2 & a_2x+b_2 & c_2 \\
    (1-x^2)a_3 & a_3x+b_3 & c_3 \\
    \end{vmatrix} = (1-x^2)\cdot \begin{vmatrix}
    a_1 & a_1x+b_1 & c_1 \\
    a_2 & a_2x+b_2 & c_2 \\
    a_3 & a_3x+b_3 & c_3 \\
    \end{vmatrix}$$
Now do the column operation $C_2-xC_1\to C_2$ to get:
$$(1-x^2)\cdot\begin{vmatrix}
    a_1 & b_1 & c_1 \\
    a_2 & b_2 & c_2 \\
    a_3 & b_3 & c_3 \\
    \end{vmatrix}.$$
