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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)`
Solution
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)`
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