by property of determinanta b-c c+ba+c b c-aa-b b+a c =(a+b+c)(a^2+b^2+c^2) Share with your friends Share 6 Manbar Singh answered this Let ∆ = ab-cc+ba+cbc-aa-ba+bcApply C1 = aC1; C2 = bC2; C3 = cC3 and divide the ∆ by abc, we get∆ = 1abca2b2-bcc2+bca2+acb2c2-aca2-abab+b2c2Applying C1 = C1 + C2 + C3, we get∆ = 1abca2+b2+c2b2-bcc2+bca2+b2+c2b2c2-aca2+b2+c2ab+b2c2⇒∆ = a2+b2+c2abc1b2-bcc2+bc1b2c2-ac1ab+b2c2Applying R1 = R1 - R2 and R2 = R2 - R3, we get⇒∆ = a2+b2+c2abc0-bcbc+ac0-ab-ac1ab+b2c2⇒∆ = a2+b2+c2abc 1abc2 + abbc+ac⇒∆ = a2+b2+c2abcabc2+ab2c+a2bc⇒∆ = a2+b2+c2abc × abcc+b+a⇒∆ =a2+b2+c2a+b+c 35 View Full Answer
