Show that
LHS = $$\begin{vmatrix}a_1+b_1t & a_2+b_2t & a_3+b_3t \\ a_1t+b_1 & a_2t+b_2 & a_3t+b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$
RHS = (1-t^2) multiply$$\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$
I will perform it from the RHS to LHS.
First, I let detA = detB = $$\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$ Then, I multiply "t" to 1st and 2nd row of detB in order to create t^2 multiply with detB. Next, I swap the 1st and 2nd row of detB to turn t^2 to -t^2. Call this detC.
Finally, I operate detA + detC in order to look identical to the LHS.
But the problem is the 3rd row of my RHS still remains $$\begin{vmatrix}a_1+b_1t & a_2+b_2t & a_3+b_3t \\ a_1t+b_1 & a_2t+b_2 & a_3t+b_3 \\2c_1 & 2c_2 & 2c_3 \\\end{vmatrix}$$
How can I eliminate those "2" in order to get my RHS = LHS ?
Thank you