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Using properties of determinants prove that: |(b + c,a - b,a),(c + a,b - c,b),(a + b,c - a,c)| = 3abc - a3 - b3 - c3

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Best answer

Considering LHS

On further calculation

= (a + b + c) [(c – a) (c + a – 2b) – (b – c) (a + b – 2c)]

By multiplication

= (a + b + c) [c2 + ca – 2bc – ca – a2 + 2ab – ab – b2 + 2bc + ac + bc – 2c2]

By simplification

= (a + b + c) [- a2 – b2 – c2 + ab + bc + ac]

Taking –ve sign as common

= – (a + b + c) [a2 + b2 + c2 + ab + bc + ac]

We get

= – (a3 + b3 + c3 – 3abc)

= 3abc – a3 – b3 – c3

= RHS

Hence, it is proved.

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