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Question Number 198862 by sonukgindia last updated on 25/Oct/23

Answered by Frix last updated on 25/Oct/23

determinant ((a,b,c),(b,c,a),(c,a,b))=3abc−(a^3 +b^3 +c^3 )= =(a+b+c)(ab+ac+bc−(a^2 +b^2 +c^2 ))= =(a+b+c)(3(ab+ac+bc)−(a+b+c)^2 )= =1×(3×0−1^2 )=−1 (x−a)(x−b)(x−c)= =x^3 −(a+b+c)_(=1) x^2 +(ab+ac+bc)_(=0) x−abc_(=−2)

|abcbcacab|=3abc−(a3+b3+c3)==(a+b+c)(ab+ac+bc−(a2+b2+c2))==(a+b+c)(3(ab+ac+bc)−(a+b+c)2)==1×(3×0−12)=−1(x−a)(x−b)(x−c)==x3−(a+b+c)=1x2+(ab+ac+bc)=0x−abc=−2

Commented by Frix last updated on 25/Oct/23

This means x^3 +px^2 +qx+r=0 determinant ((x_1 ,x_2 ,x_3 ),(x_2 ,x_3 ,x_1 ),(x_3 ,x_1 ,x_2 ))=p^3 −3pq

Thismeansx3+px2+qx+r=0|x1x2x3x2x3x1x3x1x2|=p3−3pq

Answered by cortano12 last updated on 25/Oct/23

x^3 −x^2 +2 = 0 abc = −2 ; a+b+c = 1 ab+ac+bc = 0 det(A) = 3abc−(a^3 +b^3 +c^3 ) = −6−{(a+b+c)^2 −2(ab+ac+bc)−6} = −6−{1−0−6 }= −1

x3−x2+2=0abc=−2;a+b+c=1ab+ac+bc=0det(A)=3abc−(a3+b3+c3)=−6−{(a+b+c)2−2(ab+ac+bc)−6}=−6−{1−0−6}=−1

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