Question

Prove the following identity:

`|(a^3,2,a),(b^3,2,b),(c^3,2,c)| = 2(a-b) (b-c) (c-a) (a+b+c)`

 

Answer 1

LHS 

`=|(a^3,2,a),(b^3,2,b),(c^3,2,c)|`

`=|(a^3,2,a),(b^3-a^3,0,b-a),(c^3-a^3,o,c-a)|`     `["Applying"  R_2->R_2-R_1 and R_3 -> R_3-R_1]`

`=-(a-b)(c-a)  |(a^3,2,a),(b^2+a^2+ab,0,1),(c^2+a^2+ac,o,1)|`    `["Taking "(b-a)" common from"  R_2 and  (c-a) "common from"  R_3`

`=-(a-b)(c-a) |(a^3,2,a),(b^2-c^2+ab-ac,0,0),(c^2+a^2+ac,0,1)|`    `["Applying"  R_2->R_2-R_3]`

`=-(a-b)(c-a)|(a^3,2,a),((b-c)(a+b+c),0,0),(c^2+a^2+ac,0,1)|`

`=-(a-b)(c-a)(b-c)(a+b+c)|(a^3,2,a),(1,0,0),(c^2+a^2+ac,0,1)|`    `["Taking"  (b-c)(a+b+c)   "common from"  R_2]`

`=-(a-b)(c-a)(b-c)(a+b+c)(-2)`      `["Expanding along second column"]`

`=2(a-b)(c-a)(b-c)(a+b+c)`

= RHS

∴`|(a^3,2,a),(b^3,2,b),(c^3,2,c)| = 2(a-b) (b-c) (c-a) (a+b+c)`