If the straight lines `ax+by+c=0, bx+cy+a=0` and `cx+ay+b=0` are concurrent, then prove that `a^(3)+b^(3)+c^(3)=3abc`

If the lines ax+by+c=0,bx+cy+a=0 and cx+ay+b=0 be concurrent,then:

If the lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 be concurrent, then:

The lines ax+by+c=0, bx+cy+a =0, cx+ay+b =0 are concurrent when-

If the lines ax+by+c=0, bx+cy+a=0 and Cx+ay+b=0 are concurrent (a+b+cne0) then

The three straight lines ax+by=c, bx+cy=a and cx +ay =b are collinear, if

If the three straight lines ax + 2y + 1 = 0, 3y + bx + 1 = 0 and cx+ 4y + 1=0 are concurrent, then show that- a. b, c are in AP.