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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: $$4b^4-8b^2ca+4a^3b+4bc^3-4ab^3c=0$$

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  • $\begingroup$ Recheck your calculations at the last step. $\endgroup$ – dxiv Aug 8 '17 at 16:32
  • $\begingroup$ @dxiv, Is there any alternative method for this question? $\endgroup$ – pi-π Aug 8 '17 at 16:36
  • $\begingroup$ I'd be surprised if you found anything simpler than this. $\endgroup$ – dxiv Aug 8 '17 at 19:42
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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 $0$.

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You made a mistake in the discriminant. It is

$$4b^4-8b^2ca+4a^3b+4bc3-4ab^{\color{red}{2}}c=4b\cdot(b^3-3abc+a^3+c^3)$$

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