Show that the roots of $$(a^2-bc)x^2+2(b^2-ca)x+(c^2-ab)=0$$ will be equal, if either $b=0$, or $a^3+b^3+c^3-3abc=0$.
My attempt: Comparing above equation with $Ax^2+Bx+C=0$, we get, $$A=(a^2-bc)$$ $$B=2(b^2-ca)$$ $$C=(c^2-ab)$$ According to question $$B^2-4A.C=0$$ $$[2(b^2-ca)]^2-4(a^2-bc)(c^2-ab)=0$$ On solving a few steps, I got: $$4b^4-8b^2ca+4a^3b+4bc^3-4ab^3c=0$$
Credit: Thanks fonfonx for pointing out the mistake
The discriminant $D$ is equal to $$4b^4-8b^2ca+4a^3b+4bc^3-4ab^2c$$
Let's factorize the discriminant:
$$4b(b^3-2abc+a^3+c^3-abc)$$
which is equal to
$$4b(b^3+a^3+c^3-3abc)$$
Hence if $b=0$ or $(b^3+a^3+c^3-3abc)=0$, the discriminant is $0$.
You made a mistake in the discriminant. It is
$$4b^4-8b^2ca+4a^3b+4bc3-4ab^{\color{red}{2}}c=4b\cdot(b^3-3abc+a^3+c^3)$$
