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`.

Answer 1

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`