题目 09
$$
D=\left|\begin{array}{lll}a^{2} & (a+2)^{2} & (a+4)^{2} \\ b^{2} & (b+2)^{2} & (b+4)^{2} \\ c^{2} & (c+2)^{2} & (c+4)^{2}\end{array}\right|=?
$$
解析 09
$$
D=\left|\begin{array}{lll}a^{2} & (a+2)^{2} & (a+4)^{2} \\ b^{2} & (b+2)^{2} & (b+4)^{2} \\ c^{2} & (c+2)^{2} & (c+4)^{2}\end{array}\right| \Rightarrow
$$
$$
D=\left|\begin{array}{lll}
a^{2} & a^{2}+4+4 a & a^{2}+16+8 a \\
b^{2} & b^{2}+4+4 b & b^{2}+16+8 b \\
c^{2} & c^{2}+4+4 c & c^{2}+16+8 c
\end{array}\right| \Rightarrow
$$
$$
D=\left|\begin{array}{lll}a^{2} & 4+4 a & 8 a+16 \\ b^{2} & 4+4 b & 8 b+16 \\ c^{2} & 4+4 c & 8 c+16\end{array}\right| \Rightarrow
$$
$$
D=4 \times 8\left|\begin{array}{lll}a^{2} & a+1 & a+2 \\ b^{2} & b+1 & b+2 \\ c^{2} & c+1 & c+2\end{array}\right| \Rightarrow
$$
$$
D=32\left|\begin{array}{lll}a^{2} & a+1 & 1 \\ b^{2} & b+1 & 1 \\ c^{2} & c+1 & 1\end{array}\right| \Rightarrow
$$
$$
D=32\left|\begin{array}{lll}a^{2} & a & 1 \\ b^{2} & b & 1 \\ c^{2} & c & 1\end{array}\right| \Rightarrow
$$
$$
D=32 \times(-1)^{2+1} \Rightarrow
$$
$$
\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right| \Rightarrow
$$
$$
D=-32\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right| \Rightarrow
$$
$$
D=-32(b-a)(c-a)(c-b).
$$