Prove, using the properties of determinants: determinant |(b+c)^2a^2bc (c+a)^2b^2ca (a+b)^2 c^2ab| = (a-b) (b-c) (c-a) (a + b + c)(a^2+b^2+c^2) Share with your friends Share 0 Ajanta Trivedi answered this the given determinant is: (b+c)2(c+a)2(a+b)2a2b2c2bccaab applying R1→R1-R3 and R2→R2-R3=(b+c)2-(a+b)2(c+a)2-(a+b)2(a+b)2a2-c2b2-c2c2bc-abca-abab=(c-a)(a+2b+c)-(b-c)(2a+b+c)(a+b)2-(c-a)(a+c)(b-c)(b+c)c2b(c-a)-a(b-c)ab=(b-c)(c-a)a+2b+c-2a-b-c(a+b)2-a-cb+cc2b-aab =(b-c)(c-a)-a+b-2a-b-c(a+b)2-a+b(b+c)c2-a+b-aab applying R1→R1+R2=(-a+b)(b-c)(c-a)1-2a-b-c(a+b)21(b+c)c21-aab =-(a-b)(b-c)(c-a)0-2(a+b+c)(a+b)2-c21(b+c)c21-aab applying C1→C1-C2=-(a-b)(b-c)(c-a)0-2(a+b+c)(a+b+c)(a+b-c)1(b+c)c21-aab =-(a-b)(b-c)(c-a)(a+b+c)0-2a+b-c1b+cc21-aab=-(a-b)(b-c)(c-a)(a+b+c)0-2a+b-c0b+c+ac2-ab1-aab applying C2→C2-C3now expanding the determinannts=-(a-b)(b-c)(c-a)(a+b+c)-2(c2-ab)-(a+b+c)(a+b-c)=-(a-b)(b-c)(c-a)(a+b+c)[-2c2+2ab-(a+b)2+c2]=-(a-b)(b-c)(c-a)(a+b+c)[-c2+2ab-a2-b2-2ab]=-(a-b)(b-c)(c-a)(a+b+c)[-a2-b2-c2]=(a-b)(b-c)(c-a)(a+b+c)[a2+b2+c2] = RHS hope this helps you 0 View Full Answer
