Q45 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
$\left|\begin{array}{ccc}1 & b c+a d & b^2 c^2+a^2 d^2 \\ 1 & c a+b d & c^2 a^2+b^2 d^2 \\ 1 & a b+c d & a^2 b^2+c^2 d^2\end{array}\right|=$
- A$(a-b)(b-c)(c-d)(a-d)(a-c)(d-b)$
- Correct answerB$(a-b)(a-c)(b-c)(b-d)(a-d)(c-d)$
- C$(a-b)(a-c)(a-d)(b-c)(b-d)(d-c)$
- D$(a-b)(b-c)(c-d)(b-d)$
Solution
$\left|\begin{array}{ccc}1 & 1 & 1 \\ b c+a d & c a+b d & a b+c d \\ b^2 c^2+a^2 d^2 & c^2 a^2+b^2 d^2 & a^2 b^2+c^2 d^2\end{array}\right|$
$$
\begin{aligned}
& \text { Applying } c_1 \rightarrow c_1-c_2 \text { and } c_2=c_2-c_3 \\
& \mid \begin{array}{cc}
0 & 0 \\
c(b-a)+d(a-b) & c(a-d)+b(d-a) \\
c^2\left(b^2-a^2\right)+d^2\left(a^2-b^2\right) & c^2\left(a^2-d^2\right)+b^2\left(d^2-a^2\right)
\end{array} \\
& 1 \\
& a b+c d \\
& a^2 b^2+c^2 d^2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 1\left[\{ c ( b - a ) + d ( a - b ) \} \left\{c^2\left(a^2-d^2\right)+b^2\left(d^2-a^2\right)\right.\right. \\
& \Rightarrow-\left[\{c(a-d)+b(d-a)\}\left\{c^2\left(b^2-a^2\right)+d^2\left(a^2-b^2\right)\right\}\right]
\end{aligned}
$$
Often after simplify we are getting
$$
(a-b)(a-c)(b-c)(b-d)(a-d)(c-d)
$$
Hence, the correct answer is option: 'B'.