If the roots of the equation $(c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + b^{2} - a c = 0$ are equal, prove that either $a = 0$ or $a^{3} + b^{3} = c^{3} = 3 a b c$

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Solution

We have, D = 4a $(a^{3} + b^{3} + c^{3} - 3 a b c)$
Since roots are equal.
$∴$ $D = 0 \Rightarrow 4 a (a^{3} + b^{3} + c^{3} - 3 a b c) = 0 \Rightarrow a = 0 o r, a^{3} + b^{3} + c^{3} = 3 a b c$

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