Sum of fifth roots of roots of cubic.
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$a,b,c$ are the (real) roots of $x^3-16x^2-57x+1=0$. Prove that $sqrt[5]{a} + sqrt[5]{b} +sqrt[5]{c} = 1 $ ................................................................................ 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 ==================== 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(