the three line ax+by+c=0, bx+cy+a=0, cx+ay+b=0 are concurrent only when:

(a)a+b+c=0        (b)a²+b²+c²=ab+bc+ca          (c)a³+b³+c³=3abc    (d)all of the above

the three line ax+by+c=0, bx+cy+a=0, cx+ay+b=0 are concurrent only when

just expand it to get

Hence c) is the correct answer.

Doing what versatile boy did we get,

a³+b³+c³=3abca+b+c=0 or a²+b²+c²=ab+bc+ca

If (a+b+c)=0 then (1,1) is the point of concurrence.

From the second condition,we get a=b=c which results in the three lines being same.So,I think answer is all the above.

hey allaraju plz explain the 2nd condition how did u get that?

a²+b²+c²-ab-bc-ca=(1/2)[(a-b)²+(b-c)²+(c-a)²]=0 iff a=b=c

Hence,the three eqns. represent the  same line.