A given polynomial equation of 5 degree has three equal roots . Given equation is $x^5-10a^3x^2+b^4x+c^5=0$ which has 3 equal roots. What I know is that since its a 5th degree equation therefore it must have 5 roots out (of which 3 are equal).
Aim is to establish the relationship between the constants $a,b$ and $c$.
Options given are:
1) $6a^5+c^5=0$.
2) $b^4=15a^4.$
I have to find which one is correct out of the two, any help?


Question : I have to find which one is correct out of the two, any help? 

Answer : Actually, both the options are correct. I am giving the explanation.
Let $~f(x) = x^5 – 10a^3x^2 + b^4x + c^5~$.
$$~⇒ f'(x) = 5x^4 – 20a^3x + b^4~~~~ ⇒ f'' (x) = 20x^3 – 20a^3~$$
Since $~x = \alpha~$ be a root that is repeated three times, so
$$f''(\alpha) = 0,~~~~~~ f'(\alpha) = 0,~~~~~~ f(\alpha) = 0$$
$$⇒ 20\alpha^3 – 20a^3=0,~~~~~~ 5a^4 – 20a^3\alpha + b^4 = 0,~~~~~~ \alpha^5 – 10a^3\alpha^2 + b^4\alpha +c^5 = 0$$
$$⇒ \alpha = a,~~~~~~ 5a^4 – 20a^3\alpha + b^4 = 0,~~~~~~ \alpha^5 – 10a^3\alpha^2 + b^4\alpha +c^5 = 0$$
$$⇒ b^4 = 15a^4, ~~~~~~c^5 + ab^4 – 9a^5 = 0\qquad (\text{using the  result}~~\alpha = a )~~~~$$
$$⇒ b^4 = 15a^4, ~~~~~~c^5 + 15a^5 – 9a^5 = 0\qquad (\text{using the  value}~~b^4 = 15a^4)$$
$$⇒ b^4 = 15a^4, ~~~~~~6a^5 + c^5 = 0~.$$
