Prove that if polynomial's $f(x)=x^6+ax^3+bx^2+cx+d$ , all roots are real, then $a=b=c=d=0$.

Question

$$f(x)=x^6+ax^3+bx^2+cx+d$$

I have some trouble proving it and all I know is that I should be using Vieta's formulas. Can anyone give me a hint?

Answer 1

Call $x_1 \ldots x_6$ the roots of $f$.
You have to show that if $x_1+ \ldots+ x_6 = x_1x_2+ x_1x_3 + \ldots + x_{n-2}x_n + x_{n-1}x_n = 0$,
then $x_1 = x_2 = \ldots = x_6 = 0$. Or, put another way, that $x_1^2 + x_2^2 + \ldots + x_6^2 = 0$ (because the roots are real)

Answer 2

If $f$ has six real roots then $f'$ has five, $f''$ has four, $f'''$ has three. Now $f'''(x) = 120 x^3 + 6 a$ and this only has three real roots if $a = 0$. Then look at $f''$, $f'$ and $f$.

Answer 3

You have $f(x)=0$

Let $\alpha_1, \alpha_2 \dots, \alpha_6$ be the roots of the equation.

$\sum_1^6 \alpha_k=\dfrac{0}{1}$ or simply $0$. (Why?)

$\sum_{1\le i \lt j \le 6} \alpha_i \alpha_j=0$

$(\sum_1^6 \alpha_k)^2-2\sum_{1\le i \lt j \le 6} \alpha_i \alpha_j= \sum_1^6\alpha_i^2=0 \implies \alpha_1= \cdots=\alpha_6=0$

Can you continue from here?