Prove that a^2&bc&ac+c^2 ^2+ab&b^2&ac &b^2+bc&c^2 =4a^2b^2c^2 - Vidyadip

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 C₁, C₂, and C₃, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b&b+c&c\end{vmatrix}$
Applying R₂ → R₂ + R₁ and R₃ → R₃ - R₁, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\b&b-c&-c\\b-a&b&-a\end{vmatrix}$
Applying R₂ → R₂ + R_1, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b-a&b&-a\end{vmatrix}$
Applying R₃ → R₃ + R₂, 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 R₃, we have:
$\triangle$ = 2ab²c [a(c -a) + a(a + c)]
= 2ab²c [ac - a² + a²+ ac]
= 2ab²c (2ac)
= 4a²b²c²
Hence, the given result is proved.

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