Show that if , a(b-c)x² + b(c-a)xy + c(a-b)y² is a perfect square , the quantities a,b,c are in harmonical progression.

If both x,y are 0,then the expression reduces to 0 which is a perfect square.If one of them is non-zero,say y then divide the whole expression by y² which results in a quadratic in x/y.

Hence,a(b-c)(x/y)²+b(c-a)(x/y)+c(a-b)=0,Observe that x/y=1 is a zero of the expression.So,for it to be perfect square,it must be of the form a(b-c)(x/y-1)².

Hence,-2a(b-c)=b(c-a) and a(b-c)=c(a-b).Both these result to give b=2ac/a+c which means a,b,c are in H.P.Hope you got it.

I have used the following method:

Step 1: Dividing the eq wid y²  to get  a(b-c)x²/y² + b(c-a)x/y + c(a-b) = 0

Step 2: Since for perfect square Discriminant = 0 , b² = 4 ac

            , and upon Solving, we get   ( ab+ bc - 2ac)² = 0

Step 3: This implies 2ac=b(a+c)  b= 2ac/a+c  a,b,c are in H.P