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Question:

If $f(x)=\left|\begin{array}{lll}(1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{array}\right|$ $=A+B x+C x^{2}+$_________then $A=$____________

Solution:

Let $f(x)=\left|\begin{array}{lll}(1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{array}\right|$

$f(x)=\left|\begin{array}{lll}(1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{array}\right|$

Taking out $(1+x)^{17},(1+x)^{23}$ and $(1+x)^{41}$ common from $R_{1}, R_{2}$ and $R_{3}$, respectively

$=(1+x)^{17}(1+x)^{23}(1+x)^{41}\left|\begin{array}{ccc}1 & (1+x)^{2} & (1+x)^{6} \\ 1 & (1+x)^{6} & (1+x)^{11} \\ 1 & (1+x)^{2} & (1+x)^{6}\end{array}\right|$

$=(1+x)^{81}\left|\begin{array}{ccc}1 & (1+x)^{2} & (1+x)^{6} \\ 1 & (1+x)^{6} & (1+x)^{11} \\ 1 & (1+x)^{2} & (1+x)^{6}\end{array}\right|$

Taking out $(1+x)^{2}$ and $(1+x)^{6}$ common from $C_{2}$ and $C_{3}$, respectively

$=(1+x)^{81}(1+x)^{2}(1+x)^{6}\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & (1+x)^{4} & (1+x)^{5} \\ 1 & 1 & 1\end{array}\right|$

$=(1+x)^{89}\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & (1+x)^{4} & (1+x)^{5} \\ 1 & 1 & 1\end{array}\right|$

$=(1+x)^{89}(0)$   ( $\because$ if two row $s$ are identical then the value of determinant is zero)

$=0$

Hence, $A=\underline{0}$.