1
Question
Show that ∣∣
∣∣b+ca−bac+ab−cba+bc−ac∣∣
∣∣=3abc−a3−b3−c3.
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Solution
The given determinant ∣∣
∣∣b+ca−bac+ab−cba+bc−ac∣∣
∣∣
=∣∣
∣∣baacbbacc∣∣
∣∣−∣∣
∣∣bbaccbaac∣∣
∣∣+∣∣
∣∣caaabbbcc∣∣
∣∣−∣∣
∣∣cbaacbbac∣∣
∣∣.
Of these four determinants the first three vanish, as two columns are same. Thus the expression reduces to the last of the four determinants; hence its value
=−{c(c2−ab)−b(ac−b2)+a(a2−bc)}
=3abc−a3−b3−c3.
∣∣
∣∣b+ca−bac+ab−cba+bc−ac∣∣
∣∣
=∣∣ ∣∣baacbbacc∣∣ ∣∣−∣∣ ∣∣bbaccbaac∣∣ ∣∣+∣∣ ∣∣caaabbbcc∣∣ ∣∣−∣∣ ∣∣cbaacbbac∣∣ ∣∣.
Of these four determinants the first three vanish, as two columns are same. Thus the expression reduces to the last of the four determinants; hence its value
=−{c(c2−ab)−b(ac−b2)+a(a2−bc)}
=3abc−a3−b3−c3.
=∣∣ ∣∣baacbbacc∣∣ ∣∣−∣∣ ∣∣bbaccbaac∣∣ ∣∣+∣∣ ∣∣caaabbbcc∣∣ ∣∣−∣∣ ∣∣cbaacbbac∣∣ ∣∣.
Of these four determinants the first three vanish, as two columns are same. Thus the expression reduces to the last of the four determinants; hence its value
=−{c(c2−ab)−b(ac−b2)+a(a2−bc)}
=3abc−a3−b3−c3.
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