Question
Prove that $\begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab&b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}=4a^2b^2c^2$
✓
Answer
$\triangle=\begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab&b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}$
Taking out common factors a, b, and c from $C_1, C_2,$ and $C_3$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b&b+c&c\end{vmatrix}$
Applying $R_2 → R_2 + R_1$ and $R_3 → R_3 - R_1$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\b&b-c&-c\\b-a&b&-a\end{vmatrix}$
Applying $R_2 → R_2 + R_1$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b-a&b&-a\end{vmatrix}$
Applying $R_3 → R_3 + R_2$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\2b&2b&0\end{vmatrix}$
$\triangle=2ab^2c\begin{vmatrix}a&c&a+c\\a+b&b&a\\1&1&0\end{vmatrix}$
$\triangle=2ab^2c\begin{vmatrix}a&c-a&a+c\\a+b&-a&a\\1&0&0\end{vmatrix}$
Expanding along $R_3$, we have:
$\triangle = 2ab^2c [a(c -a) + a(a + c)]$
$= 2ab^2c [ac - a^2 + a^2+ ac]$
$= 2ab^2c (2ac)$
$= 4a^2b^2c^2$
Hence, the given result is proved.
Taking out common factors a, b, and c from $C_1, C_2,$ and $C_3$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b&b+c&c\end{vmatrix}$
Applying $R_2 → R_2 + R_1$ and $R_3 → R_3 - R_1$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\b&b-c&-c\\b-a&b&-a\end{vmatrix}$
Applying $R_2 → R_2 + R_1$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b-a&b&-a\end{vmatrix}$
Applying $R_3 → R_3 + R_2$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\2b&2b&0\end{vmatrix}$
$\triangle=2ab^2c\begin{vmatrix}a&c&a+c\\a+b&b&a\\1&1&0\end{vmatrix}$
$\triangle=2ab^2c\begin{vmatrix}a&c-a&a+c\\a+b&-a&a\\1&0&0\end{vmatrix}$
Expanding along $R_3$, we have:
$\triangle = 2ab^2c [a(c -a) + a(a + c)]$
$= 2ab^2c [ac - a^2 + a^2+ ac]$
$= 2ab^2c (2ac)$
$= 4a^2b^2c^2$
Hence, the given result is proved.
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