Prove that: `|[a, b, ax+by],[ b, c, bx+cy], [ax+by, bx+cy,0]|=(b^2-a c)(a x^2+2b x y+c y^2)`

Prove that |(a,b,ax+by),(b,c,bx+cy),(ax+by, bx + cy, 0)| = (b^(2)-ac)(ax^(2) + 2bxy + cy^(2)) .

|{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}|=(b^2-ac)(ax^2+2bxy+cy^2)

det[[ Prove that: ,b,ax+bya,c,bx+cyax+by,bx+cy,0]]=(b^(2)-ac)(ax^(2)+2bxy+cy^(2))

[[a, b, ax + byb, c, bx + cyax + by, bx + cy, 0]] = (b ^ (2) -ac) (ax ^ (2) + 2bxy + cy ^ (2))

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then Delta = |(a,b,ax +b),(b,c,bx +c),(ax +b,bx +c,0)| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If |{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}|=0 and ax^2+2abxy+cy^2ne0," then "......