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

(c²−ab)x²−(a²−bc)x+(b²−ac)=0
comparing with Ax²+Bx+C=0, we get
∴A=(c²−ab),B=−(a²−bc),C=(b²−ac)
Roots of the quadratic eqn. are equal ⇒D=0⇒√B²−4AC=0⇒B²−4AC=0⇒B²=4AC⇒[−(a²−bc)]²=4(c²−ab)(b²−ac)
⇒4(a²−bc)²=4(c²−ab)(b²−ac)
⇒a⁴+b²c²−2a²bc=b²c²−ac³−ab³+a²bc
⇒a⁴+ab³+ac³=3a²bc.
⇒a(a³+b³+c³−3abc)=0.
∴a=0 or a³+b³+c³=3abc.