Viete's Relations: Solving Cubic Equations

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I'm having trouble with number three. I know Viete's relations are X1+X2+X3, X1X2+X1X3+X2X3, and x1x2x3 for a cubic equation.
 

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A symmetric polynomial doesn't change after any permutation.
 
correct, but how do i show that?
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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