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Question

# Show that:∣∣ ∣∣abcbcacab∣∣ ∣∣2=∣∣ ∣ ∣∣2bc−a2c2b2c22ac−b2a2b2a22ab−c2∣∣ ∣ ∣∣=(a3+b3+c3−3abc)2

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Solution

## ∣∣ ∣∣abcbcacab∣∣ ∣∣=−∣∣ ∣∣acbbaccba∣∣ ∣∣=∣∣ ∣∣−acb−bac−cba∣∣ ∣∣So ∣∣ ∣∣abcbcacab∣∣ ∣∣⋅∣∣ ∣∣abcbcacab∣∣ ∣∣=∣∣ ∣∣−acb−bac−cba∣∣ ∣∣⋅∣∣ ∣∣abcbcacab∣∣ ∣∣As detA.detB=det(AB) for two matricesHere AB=⎡⎢⎣−acb−bac−cba⎤⎥⎦⋅⎡⎢⎣abcbcacab⎤⎥⎦=⎡⎢ ⎢⎣−a.a+bc+bc−ab+c2+ab−ac+ac+b2−ba+ba+c2−b2+ac+ac−bc+c2+bc−ca+b2+ac−bc+bc+c2−c2+ab+ab⎤⎥ ⎥⎦=⎡⎢ ⎢⎣2bc−a2c2b2c22ac−b2a2b2a22ab−c2⎤⎥ ⎥⎦So |AB|=|A|.|B|\$ie ∣∣ ∣ ∣∣2bc−a2c2b2c22ac−b2a2b2a22ab−c2∣∣ ∣ ∣∣=∣∣ ∣∣abcbcacab∣∣ ∣∣∣∣ ∣∣−acb−bac−cba∣∣ ∣∣∣∣ ∣ ∣∣2bc−a2c2b2c22ac−b2a2b2a22ab−c2∣∣ ∣ ∣∣=∣∣ ∣∣abcbcacab∣∣ ∣∣∣∣ ∣∣abcbcacab∣∣ ∣∣∣∣ ∣ ∣∣2bc−a2c2b2c22ac−b2a2b2a22ab−c2∣∣ ∣ ∣∣=∣∣ ∣∣abcbcacab∣∣ ∣∣2∣∣ ∣∣abcbcacab∣∣ ∣∣=a(bc−a2)−b(b2−ac)+c(ab−c2)∣∣ ∣∣abcbcacab∣∣ ∣∣=3abc−a3−b3−c3∣∣ ∣∣abcbcacab∣∣ ∣∣2=(a3+b3+c3−3abc)2∣∣ ∣ ∣∣2bc−a2c2b2c22ac−b2a2b2a22ab−c2∣∣ ∣ ∣∣=∣∣ ∣∣abcbcacab∣∣ ∣∣2=(a3+b3+c3−3abc)2

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