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 May 11th, 2013, 06:08 AM #1 Senior Member   Joined: Aug 2011 Posts: 149 Thanks: 0 Need help with proving Hello. Could anyone help me to prove that: Prove that when zero points of polynomial$f(x)= x^6 + ax^3 + bx^2+cx +d$ are real then $a= b = c = d = 0$. Hint: vieta's formula. Here's vieta's formule i've written. Now how do I show that a= b= c = d = 0? $\begin{cases} x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 0\\ x_1x_2 + x_1x_3 + x_1x_4 + x_1x_5 + x_1x_6 + x_2x_3 + x_2x_4 + x_2x_5 + x_2x_6 + x_3x_4 + x_3x_5 + x_3x_6 + x_4x_5 + x_4x_6 + x_5x_6 = 0\\ x_1x_2x_3 + x_1x_2x_4 + x_1x_2x_5 + x_1x_2x_6 + x_1x_3x_4 + x_1x_3x_5 + x_1x_3x_6 + x_1x_4x_5 + x_1x_4x_6 + x_1x_5x_6 +\\ \;\;\;\;\; x_2x_3x_4 + x_2x_3x_5 + x_2x_3x_6 + x_2x_4x_5 + x_2x_4x_6 + x_2x_5x_6 + x_3x_4x_5 + x_3x_4x_6 + x_3x_5x_6 + x_4x_5x_6 = -a\\ x_1x_2x_3x_4 + x_1x_2x_3x_5 + x_1x_2x_3x_6 + x_1x_2x_4x_5 + x_1x_2x_4x_6 + x_1x_2x_5x_6 + x_1x_3x_4x_5 + x_1x_3x_4x_6 +\\ \;\;\;\; x_1x_3x_5x_6 + x_1x_4x_5x_6 + x_2x_3x_4x_5 + x_2x_3x_4x_6 + x_2x_3x_5x_6 + x_2x_4x_5x_6 + x_3x_4x_5x_6 = b\\ x_1x_2x_3x_4x_5 + x_1x_2x_3x_4x_6 + x_1x_2x_3x_5x_6 + x_1x_2x_4x_5x_6 + x_1x_3x_4x_5x_6 + x_2x_3x_4x_5x_6 = -c\\ x_1x_2x_3x_4x_5x_6 = d \end{cases}$ Edit: I found that if x1..x6 were 0 then it works. fut it that formal enough?
 May 11th, 2013, 08:43 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Need help with proving Not quite. If you can show that the only real solution is $(x_1, x_2, x_3, x_4, x_5, x_6)= (0, 0, 0, 0, 0, 0)$ then you would have proven that $a= b = c = d = 0$. You have shown that $(0, 0, 0, 0, 0, 0)$ is a solution, but you have not shown that it is the only solution.
 May 11th, 2013, 09:29 AM #3 Senior Member   Joined: Aug 2011 Posts: 149 Thanks: 0 Re: Need help with proving how would I show that?
 May 11th, 2013, 09:49 AM #4 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Need help with proving Here is one way: $x_1 + x_2 + x_3 + x_4 + x_5 + x_6= 0$ Hence, $(x_1 + x_2 + x_3 + x_4 + x_5 + x_6)^2= 0$ $x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + 2(x_1x_2 + x_1x_3 + x_1x_4 + x_1x_5 + x_1x_6 + x_2x_3 + x_2x_4 + x_2x_5 + x_2x_6 + x_3x_4 + x_3x_5 + x_3x_6 + x_4x_5 + x_4x_6 + x_5x_6)= 0$ $x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + 2(0)= 0$ $x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2= 0$ If all of the x's are real, then their squares must be nonnegative, and hence the only way for the sum of their squares to be equal to 0 is for all of them to be equal to 0.
 May 11th, 2013, 10:15 AM #5 Senior Member   Joined: Aug 2011 Posts: 149 Thanks: 0 Re: Need help with proving Thank you for your response.

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