Determinant Prove

Asked by thebluebloo | 28th May, 2009, 04:59: PM

Expert Answer:

opening the determinent

a(bc-(b+a)(c-a))-(b-c)(c(c+a)-(a-b)(c-a))+(b+c)((a+c)(b+c)-(a-b)b)

=a(bc-bc+ba-ac+a2)-(b-c)(c2+ca-ac+a2+bc-ba)+(b+c)(ab+ac+bc+c2-ab+b2)

=a(ba-ac+a2)-(b-c)(c2+a2+bc-ab)+(b+c)(ac+bc+c2+b2)

=a2b-a2c+a3-bc2-ba2-b2c+ab2+c3+a2c+bc2-abc+abc+b2c+bc2+b3+ac2+bc2+c3+b2c

=a3+ab2+ac2+ba2+b3+bc2+a2c+b2c+c3

 =(a+b+c)(a2+b2+c2))

Answered by  | 28th May, 2009, 06:58: PM

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