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?
 A: Let $$f(x) = x^5 - 10a^3x^2 + b^4x + c^5 = (x - m)^3g(x)$$ where $m$ is the repeated root, and $g(x)$ is some second-order polynomial. 
Then, differentiating and substituting $x = m$,
$$5x^4 - 20a^3x + b^4 = (x - m)^3g'(x) + 3(x - m)^2g(x)$$
$$5m^4 - 20a^3m + b^4 = 0$$
Differentiating another time, and similarly substituting $x = m$,
$$20x^3 - 20a^3  = (x - m)^3g''(x) + 3(x - m)^2g'(x) + 3(x - m)^2g'(x) + 6(x - m)g(x)$$
$$20m^3 - 20a^3 = 0$$
$$m = a$$
Substituting this back into our first result,
$$5a^4 - 20a^3\cdot a + b^4 = 0$$
$$b^4 = 15a^4$$

Note: The fact that $(x - m)$ appears in every term of $f'(x)$ and $f''(x)$ is in fact a usable basis for proving what Gerry Myerson mentioned in the comments - that $f(x), f'(x), f''(x)$ all have a common factor.
A: Let $t$ the triple root and consider the coefficients $r,s$ of the quadratic factor so one has the polynomial $(x-t)^3(x^2+rx+s)$. The equality of corresponding coefficients gives $$r-3t=0$$ $$s-3tr+3t^2=0$$ $$-3ts-t^3+3t^r=-10a^3$$  $$3t^2s-t^3r=b^4$$  $$-t^3s=c^5$$   Solving easily this system we have a parameterization of $a,b,c$ in function of the triple root.
$$a=t$$ $$b=\sqrt[4]{15}\space t$$   $$c=-\sqrt[5]{6}\space t$$
The simplest relationship between $a,b,c$ is $$a=\frac{b}{\sqrt[4]15}=\frac{-c}{\sqrt[5]6}$$
A: 
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
$$f''(\alpha) = 0,~~~~~~ f'(\alpha) = 0,~~~~~~ f(\alpha) = 0$$
$$⇒ 20\alpha^3 – 20a^3=0,~~~~~~ 5a^4 – 20a^3\alpha + b^4 = 0,~~~~~~ \alpha^5 – 10a^3\alpha^2 + b^4\alpha +c^5 = 0$$
$$⇒ \alpha = a,~~~~~~ 5a^4 – 20a^3\alpha + b^4 = 0,~~~~~~ \alpha^5 – 10a^3\alpha^2 + b^4\alpha +c^5 = 0$$
$$⇒ b^4 = 15a^4, ~~~~~~c^5 + ab^4 – 9a^5 = 0\qquad (\text{using the  result}~~\alpha = a )~~~~$$
$$⇒ b^4 = 15a^4, ~~~~~~c^5 + 15a^5 – 9a^5 = 0\qquad (\text{using the  value}~~b^4 = 15a^4)$$
$$⇒ b^4 = 15a^4, ~~~~~~6a^5 + c^5 = 0~.$$
