If the equation $x^{5} - 10 a^{3} x^{2} + b^{4} x + c^{5} = 0$ has three equal roots, show that $a b^{4} - 9 a^{5} + c^{5} = 0$ .

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Solution

Let
$f (x) = x^{5} - 10 a^{3} x^{2} + b^{4} x + c^{5} = (x - m)^{3} g (x)$
where $m$ is the repeated root,
and $g (x)$ is some second-order polynomial.
Then, differentiating and substituting $x = m$ ,
$5 x^{4} - 20 a^{3} x + b^{4} = (x - m)^{3} g^{'} (x) + 3 (x - m)^{2} g (x)$
$5 m^{4} - 20 a^{3} m + b^{4} = 0 \dots e q n (1)$
Differentiating another time, and similarly substituting, $x = m$
$20 m^{3} - 20 a^{3} = 0$
$m = a$
Substituting this back into our equation
$a^{5} - 10 a^{5} + b^{4} a + c^{5} = 0 \Rightarrow a b^{4} - 9 a^{5} + c^{5} = 0$

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