Question

If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc

Answer 1

(c2-ab)x2 - 2(a2-bc)x+(b2 - ac) = 0
for equal roots b2=4ac
ie 4(a2-bc)2=4 (c2-ab).(b2-ac)
4 cancel from both side
a4 +b2c2 -2a2bc = b2c2-ac3-ab3 + a2bc
b2c2 cancel here
a4+ac3+ab3 =3a2bc
a(a3 +b3 +c3)=3a2bc
here a cancel from both sides
a3+b3+c3=3abc proved