ntProve the determinant a b-c c-bn nt a-c b c-a = (a+b-c)(b+c-a)(c+a-b)n nt a-b b-a c n

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∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b - a & 1 & c - b & - c & 1 \end{matrix} ∣ ∣ ∣ ∣ =

X^{\frac{a + b - c}{(a - c) (b - c)}} . X^{\frac{b + c - a}{(b - a) (c - a)}} . X^{\frac{c + a - b}{(c - b) (a - b)}} =

$4 B + 3 O_{2} \to A$

$2 B + N_{2} \to B$

$A & B$ are as follows

(a + b) (a - b) + (b + c) (b - c) + (c + a) (c - a)

(a + b) (a - b) + (b + c) (b - c) + (c + a) (c - a)

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