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Question
Using properties of determinants, prove the following:
`|(a, b,c),(a-b, b-c, c-a),(b+c, c+a, a+b)| = a^3 + b^3 + c^3 - 3abc`.
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Solution
Δ = `|(a, b,c),(a-b, b-c, c-a),(b+c, c+a, a+b)|`
= `|(a,b,c),(a-b, b-c, c-a),(a+b+c, c+a+b, a+b+c)| ...[ "Applying" R_3 -> R_3 + R_1]`
= `(a +b+c) |(a,b,c),(a-b, b-c, c-a),(1,1,1)| ...["Taking" (a +b +c) "common"]`
= `(a +b+c) |(a, b ,c),(-b , -c, -a),(1, 1, 1)| ...["Applying" R_2 -> R_2 - R_1]`
= `(a +b+c) |(a-c, b-c, c),(-b +a , -c+a, -a),(0, 0, 1)| ...[C_1 -> C_1 - C_3 "and" C_2 ->C_2 - C_3]`
= `(a +b+c) [{(a-c) (a-c) -(b-c)(a-b)}]`
= `(a +b+c) ...[{(a - c)^2 - (ab - ac - b^2 + bc)]`
= `(a +b+c) [(a -c ^2- (ab - ac - b^2 + bc)]`
= `(a +b+c) (a^2 + b^2 + c^2 - ab - bc - ca)`
= `a^3 + b^3 + c^3 - 3abc`
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