If the equation $$x^{5} - 10a^{3}x^{2} + b^{4}x + c^{5} = 0$$ has three equal roots, show that $$ab^{4} - 9a^{5} + c^{5} = 0$$.
Let
$$f(x)=x^5-10a^3x^2+b^4x+c^5=(x-m)^3g(x)$$
f(x)=x5−10a3x2+b4x+c5=(x−m)3g(x)
where $$m$$ is the repeated root,
and $$g(x)$$is some second-order polynomial.
g(x
Then, differentiating and substituting $$x=m$$,
$$5x^4-20a^3x+b^4=(x-m)^3g'(x)+3(x-m)^2g(x)$$
$$5m^4-20a^3m+b^4=0\dots eqn(1)$$
5m4−20a3m+b4=0
Differentiating another time, and similarly substituting,$$x=m$$
$$20m^3-20a^3=0$$
$$m=a$$
Substituting this back into our equation
$$a^5-10a^5+b^4a+c^5=0\\ \Rightarrow ab^4-9a^5+c^5=0$$
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