Prove that: `|a b a x+b y b c b x+c y a x+b y b x+c y0|=(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-a c)(a x^2+2b x y+c y^2)

Prove that |a x-b y-c z a y+b x c x+a z a y+b x b y-c z-a x b z+c y c x+a z b z+c y c z-a x-b y|=(x^2+y^2+z^2)(a^2+b^2+c^2)(a x+b y+c z)dot

Prove that (a b+x y)(a x+b y)>4a b x y(a , b ,x ,y >0)dot

Prove that (a b+x y)(a x+b y)>4a b x y(a , b ,x ,y >0)dot

Show that |a b c a+2x b+2y c+2z x y z|=0

Prove that |[a x-b y-c z, a y+b x, c x+a z], [a y+b x, b y-c z-a x, b z+c y],[c x+a z, b z+c y, c z-a x-b y]|=(x^2+y^2+z^2)(a^2+b^2+c^2)(a x+b y+c z)dot

If x+y+z=0 prove that |[a x, b y, c z],[ c y, a z ,b x],[ b z ,c x ,a y]|=x y z|[a, b, c],[c ,a ,b],[b ,c ,a]|

Let a ,b ,c be real numbers with a^2+b^2+c^2=1. Show that the equation |a x-b y-c b x-a y c x+a b x+a y-a x+b y-cc y+b c x+a c y+b-a x-b y+c|=0 represents a straight line.

The determinant |y^2-x y x^2a b c a ' b ' c '| is equal to a. |b x+a y c x+b y b^(prime)x+a ' y c^(prime)x+b ' y| b. |a x+b y b x+c y a^(prime)x+b ' y b ' x+c ' y| c. |b x+c y a x+b y b^(prime)x+c ' y a^(prime)x+b ' y| d. |a x+b y b x+c y a^(prime)x+b ' y b^(prime)x+c ' y|

Solve: a x+b y=c ,\ \ \ \ b x+a y=1+c