If the roots ff 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"

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Balaji Puppala

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Compare given equation with px^2+qx+r=0 We get p=c^2-ab, q=-2(a^2-bc), r=b^2-ac Given roots are equal discriminate q^2-4pr=0 4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0 a^4+b^2c^2-2a^2bc-c^2b^2+ab^3+ac^3-a^2bc=0 a^4-3a^2bc+ab^3+ac^3=0 a(a^3+b^3+c^3-3abc)=0 a=0, a^3+b^3+c^3-3abc=0 a=0, a^3+b^3+c^3=3abc
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