TITLE: Sum of fifth roots of roots of cubic.
QUESTION [10 upvotes]: $a,b,c$ are the (real) roots of $x^3-16x^2-57x+1=0$. Prove that
$\sqrt[5]{a} + \sqrt[5]{b} +\sqrt[5]{c} = 1 $
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edit : my answer to this question on another forum
Let $p=\sqrt[5]{a},q=\sqrt[5]{b}, r=\sqrt[5]{c}, u=p+q+r, v=qr+rp+pq, w=pqr$ 
From given cubic p⁵+q⁵+r⁵=16, q⁵r⁵+r⁵p⁵+p⁵q⁵=-57, p⁵q⁵r⁵=-1 (so w=-1)
Thus results at end gives the following simultaneous equations for u,v
p⁵+q⁵+r⁵ = 16 = u⁵-5u³v+5uv²-5u²+5v 
q⁵r⁵+r⁵p⁵+p⁵q⁵ = -57 = v⁵+5uv³+5u²v+5v²+5u 
I solved these via Mathematica to give u=1, v=-2 (only real solution) so your answer is 1 
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p,q,r are roots of z³-uz²+vz-w=0
Multiply by zⁿ and let s(n)=pⁿ+qⁿ+rⁿ 
Set z=p,q,r in turn and sum for the recurrence s(n+3) = us(n+2)-vs(n+1)+ws(n) 
Noting that s(0)=3, s(1)=u, s(2)=u²−2v generate s(3), s(4), s(5) as follows …
n=0 : s(3) = us(2)-vs(1)+ws(0) = u³-3uv+3w
n=1 : s(4) = us(3)-vs(2)+ws(1) = u⁴-4u²v+2v²+4uw
n=2 : s(5) = us(4)-vs(3)+ws(2) = u⁵-5u³v+5uv²+5u²w-5vw 
p⁵+q⁵+r⁵ = u⁵-5u³v+5uv²+5u²w-5vw 
Replace p by qr, q by rp, r by pq and get
q⁵r⁵+r⁵p⁵+p⁵q⁵ = v⁵-5uv³w+5u²vw²+5v²w²-5uw³ 
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REPLY [3 votes]: The given polynomial $p(x)$ has three real roots so $p(x^5)$ also has three real roots as all others are complex fifth roots of reals.
So we need to factor $p(x^5)=q(x)r(x)$ where $q$ is cubic with real roots and $r$ is an unimportant polynomial of degree twelve with complex ones.
With a simple numerical solver or with Cardano's formulas, we estimate the coefficients of $q$ using Vieta's formulas on the fifth roots of $p$, giving
$$q(x)\approx (x-1.80194) (x-0.445042) (x+1.24698)\\\approx x^3-1.000002 x^2-2.000002633x+1.000001871.$$
So we can hypothetize $q(x)=x^3-x^2-2x+1$ and long division confirms this hypothesis exactly *.
Then the sum of the real roots of $q$ is $1$.

*$$\dfrac{x^{15}-16x^{10}-57x^5+1}{x^3-x^2-2x+1}\\=x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1.$$