* Problem: *If the roots of the equation \[{(x-a)(x-b)}+{(x-b)(x-c)}+{(x-c)(x-a)}=0\],(where a,b,c are real numbers) are equal , then

**(A) \[b^2-4ac=0\]**

**(B) \[a=b=c\]**

**(C) ****\[a+b+c=0\]**

**(D)**** none of foregoing statements is correct**

**Answer: **ans (B)

\[{(x-a)(x-b)}+{(x-b)(x-c)}+{(x-c)(x-a)}=0\]

\[=>x^2-{(a+b)}x+ab+x^2-{(b+c)}x+bc+x^2-{(c+a)}x+ca=0\]

\[=>3x^2-2{(a+b+c)}x+(ab+bc+ca)=0\]

discriminant, of the equation is

\[4{(a+b+c)^2}-4.3{(ab+bc+ca)}=o\]

\[=>a^2+b^2+c^2+2(ab+bc+ca)-3(ab+bc+ca)=0\]

\[=>a^2+b^2+c^2-(ab+bc+ca)=0\]

\[=>a=b=c\]

So, option * (B) *is correct ….