Vieta's Relations: Proving \sumg(x_{k}) = 6

[muzak]

Homework Statement

Consider the polynomials:

f(x) = x[itex]^{6}[/itex] + x[itex]^{3}[/itex] +1 and g(x) = x[itex]^{2}[/itex] + x + 1

Denote the roots of f(x) = 0 by x[itex]_{1}[/itex], ... , x[itex]_{6}[/itex].

Show that [itex]\sum[/itex]g(x[itex]_{k}[/itex]) = 6 , 1[itex]\leq[/itex]k[itex]\leq[/itex]6

Homework Equations

Vieta relations.

The Attempt at a Solution

Please correct any initial mistakes I may have made in this, new to Vieta.

[itex]\sum[/itex]x[itex]_{i}[/itex] = 0, 1[itex]\leq[/itex]i[itex]\leq[/itex]6

[itex]\sum[/itex]x[itex]_{i}[/itex]x[itex]_{j}[/itex]x[itex]_{k}[/itex] = -1, 1[itex]\leq[/itex]i,j,k[itex]\leq[/itex]6, i[itex]\neq[/itex]j[itex]\neq[/itex]k

and

[itex]\prod[/itex]x[itex]_{i}[/itex] = 1, 1[itex]\leq[/itex]i[itex]\leq[/itex]6.

Then we must show that

x[itex]^{2}_{1}[/itex] + x[itex]_{1}[/itex] + 1 + x[itex]^{2}_{2}[/itex] + x[itex]_{2}[/itex] + 1 + x[itex]^{2}_{3}[/itex] + x[itex]_{3}[/itex] + 1 + x[itex]^{2}_{4}[/itex] + x[itex]_{4}[/itex] + 1 + x[itex]^{2}_{5}[/itex] + x[itex]_{5}[/itex] + 1 + x[itex]^{2}_{6}[/itex] + x[itex]_{6}[/itex] + 1 = 6

or using the first vieta relation we must show that

x[itex]^{2}_{1}[/itex] + x[itex]^{2}_{2}[/itex] + x[itex]^{2}_{3}[/itex] + x[itex]^{2}_{4}[/itex] + x[itex]^{2}_{5}[/itex] + x[itex]^{2}_{6}[/itex] = 0

Then I'm stuck, if I did everything right so far.

Never mind, I think I might have gotten it, I could just multiply the first vieta relation by itself to get what I need, the x's square, which will include the second two vieta relations which'll add up to zero and all the other combinations of the x's will automatically be zero since we had no coefficients for some of the polynomial degrees, right?

 

[Dick]

Right, I think. You didn't really show the whole thing. But sure, if you square the first relation you'll get the sum of the squares plus something else, which you can also show to be zero.