# Thread: Need help with proving.

1. ## [solved] Need help with proving.

Hello.

Could anyone help me to prove that:
Prove that when zero points of polynomial$\displaystyle f(x) = x^6 + ax^3 + bx^2+cx +d$ are real then $\displaystyle a = b = c = d = 0$.
Hint: vieta's formula.

Here's vieta's formule i've written. Now how do I show that a= b= c = d = 0?
$\displaystyle f(x) = x^6 + ax^3 + bx^2+cx +d = f(x) = x^6 + 0x^5 + 0x^4 + ax^3 + bx^2+cx +d$

$\displaystyle \begin{cases}x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 0\\x_1x_2 + x_1x_3 + x_1x_4 + x_1x_5 + x_1x_6 + x_2x_3 + x_2x_4 + x_2x_5 + x_2x_6 + x_3x_4 + x_3x_5 + x_3x_6 + x_4x_5 + x_4x_6 + x_5x_6 = 0\\x_1x_2x_3 + x_1x_2x_4 + x_1x_2x_5 + x_1x_2x_6 + x_1x_3x_4 + x_1x_3x_5 + x_1x_3x_6 + x_1x_4x_5 + x_1x_4x_6 + x_1x_5x_6 +\\ \;\;\;\;\; x_2x_3x_4 + x_2x_3x_5 + x_2x_3x_6 + x_2x_4x_5 + x_2x_4x_6 + x_2x_5x_6 + x_3x_4x_5 + x_3x_4x_6 + x_3x_5x_6 + x_4x_5x_6 = -a\\x_1x_2x_3x_4 + x_1x_2x_3x_5 + x_1x_2x_3x_6 + x_1x_2x_4x_5 + x_1x_2x_4x_6 + x_1x_2x_5x_6 + x_1x_3x_4x_5 + x_1x_3x_4x_6 +\\ \;\;\;\; x_1x_3x_5x_6 + x_1x_4x_5x_6 + x_2x_3x_4x_5 + x_2x_3x_4x_6 + x_2x_3x_5x_6 + x_2x_4x_5x_6 + x_3x_4x_5x_6 = b\\x_1x_2x_3x_4x_5 + x_1x_2x_3x_4x_6 + x_1x_2x_3x_5x_6 + x_1x_2x_4x_5x_6 + x_1x_3x_4x_5x_6 + x_2x_3x_4x_5x_6 = -c\\x_1x_2x_3x_4x_5x_6 = d\end{cases}$

Edit:I figured that if I take x1..x6 = 0 then $\displaystyle a = b = c = d = 0$.