Question

If the roots of the equation $$(c^{2}-ab)x^{2}-2(a^{2}-bc)x+(b^{2}-ac)=0$$ are equal. Prove that either $$a=0$$ or $$a^{3}+b^{3}+c^{3}=3abc$$.

Solution

$$b^{2}-4ac=0$$$$\left [ -2(a^{2}-bc) \right ]^{2}-4(c^{2}-ab)(b^{2}-ac)=0$$$$4a(a^{3}+b^{3}+c^{3}-3abc)=0$$$$\therefore a=0\ or\ a^{3}+b^{3}+c^{3}=3abc$$Mathematics

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