Sum of fifth roots of roots of cubic.

Question

$a,b,c$ 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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Answer 1

Let $A,B,C$ be the roots of the cubic polynomial $$ g(x) = x^3 - x^2 - 2x + 1\ . $$ So $A,B,C$ satisfy (Vieta): $$ \begin{aligned} e_1 &=A + B + C &&=+1\ ,\\ e_2 &= AB + BC + CA &&=-2\ ,\\ e_3 &= A B C &&= -1\ . \end{aligned} $$ (Computer support motivating the above is postponed.) These values are numerically:

sage: R.<x> = PolynomialRing(QQ)
sage: g = x^3 - x^2 - 2*x + 1
sage: g(x).roots(ring=QQbar, multiplicities=False)
[-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?]

so they are real. We compute the polynomial with roots $a'=A^5$, $b'=B^5$, $c'=C^3$.

(We "hope" that $a=a'$, $b=b'$, $c=c'$.)

For this we have to start computations involving symmetrical polynomials, trying to get $e_1(a',b',c')$, $e_2(a',b',c')$, $e_3(a',b',c')$, in terms of $e_1(A,B,C)$, $e_2(A,B,C)$, $e_3(A,B,C)$. Here, $e_1$, $e_2$, $e_3$ are the first three symmetric polynomials. (We have formally $e_4=e_5=\dots=0$.)

One result is immediate: $$ e_3(a',b',c') =a'b'c' =A^5B^5C^5=(ABC)^5=e_3(A,B,C)^5=-1\ . $$ We need now the Newton polynomial $p_5$ in Newton's identities of fifth degree... $$ \begin{aligned} e_1(a',b',c') &= a'+b'+c' \\ &=A^5+B^5+C^5 \\ &=p_5(A,B,C)\ ,\\ &\qquad\text{so we compute successively}\\ p_1(A,B,C) &= e_1(A,B,C)=1\ ,\\ p_2(A,B,C) &= (e_1p_1-2e_2)(A,B,C)=1\cdot 1-2\cdot(-2)=5\ ,\\ p_3(A,B,C) &= (e_1p_2-e_2p_1+3e_2)(A,B,C)=1\cdot 5-(-2)\cdot1+3\cdot(-1)=4\ ,\\ p_4(A,B,C) &= (e_1p_3-e_2p_2+e_3p_1)(A,B,C)=1\cdot 4-(-2)\cdot5+(-1)\cdot1=13\ ,\\ e_1(a',b',c') &=p_5(A,B,C) \\&=(e_1p_4-e_2p_3+e_3p_2)(A,B,C)=1\cdot13-(-2)\cdot4+(-1)\cdot5=16\ . \end{aligned} $$ We need now finally $a'b'+b'c'+c'a'$. For this, we repeat the same procedure as above, but not for $A,B,C$, but for $s,t,u$, which are respectively $AB$, $BC$, $CA$, with $$ \begin{aligned} e_1(s,t,u) &=s+t+u=AB + BC + CA &&=-2\ ,\\ e_2(s,t,u) &=st+tu+us= ABC(A+B+C) &&=-1\ ,\\ e_3(s,t,u) &=stu= A^2 B^2 C^2 &&= +1\ . \end{aligned} $$ So we have: $$ \begin{aligned} p_1(s,t,u)&=e_1(s,t,u)=-2\ ,\\ p_2(s,t,u)&=(e_1p_1-2e_2)(s,t,u)=(-2)\cdot (-2)-2\cdot(-1)=6\ ,\\ p_3(s,t,u)&=(e_1p_2-e_2p_1+3e_2)(s,t,u)=(-2)\cdot6-(-1)\cdot(-2)+3\cdot1=-11\ ,\\ p_4(s,t,u)&=(e_1p_3-e_2p_2+e_3p_1)(s,t,u) =(-2)\cdot(-11)-(-1)\cdot6+1\cdot(-2)=26\ ,\\ e_2(a',b',c')&=p_5(s,t,u) \\ &=(e_1p_4-e_2p_3+e_3p_2)(s,t,u) =(-2)\cdot26-(-1)\cdot(-11)+1\cdot6=-57\ . \end{aligned} $$ So $a',b',c'$ are the roots of the given polynomial $$ x^3-E_1x^2+E_2x-E_3 = x^3 - 16x^2 -57 x +1\ . $$ So $a',b',c'$ are (up to reordering) the "given" values $a,b,c$. We conclude: $$ a^{1/5} + b^{1/5} + c^{1/5} =A+B+C=1\ . $$

$\square$

Numerical support for the computations done so far, and the motivation for the abrupt start with the roots $A,B,C$ of the polynomial $x^3 - x^2 - 2x + 1$:

Sage code:

R.<x> = PolynomialRing(QQ)
f = x^3  - 16*x^2 - 57*x + 1
a, b, c = f.roots( ring=QQbar, multiplicities=False )
a, b, c
a^(1/5), b^(1/5), c^(1/5)

This gives:

(-3.015065490237851?, 0.01745839634379104?, 18.99760709389406?)
(1.008827691046369? + 0.7329562209746396?*I,
 0.4450418679126288?,
 1.801937735804839?)

OK, the computer needs human assitance to get the right fifth root of $a$ in $\Bbb R$.

 A, B, C = -(-a)^(1/5), b^(1/5), c^(1/5)

giving

sage: A, B, C
(-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?)

this is good, three real numbers. Let us compute the elementary symmetric functions, and the first Newton polynomials for them:

print "A + B + C    = %s" % (A+B+C)
print "AB + BC + CA = %s" % (A*B + B*C + C*A) 
print "A B C        = %s" % (A*B*C)

for k in [1..5]:
    print "A^%s + B^%s + C^%s = %s" % (k, k, k, A^k+B^k+C^k)

This gives so far:

A + B + C    = 1.000000000000000?
AB + BC + CA = -2.000000000000000?
A B C        = -1.000000000000000?
A^1 + B^1 + C^1 = 1.000000000000000?
A^2 + B^2 + C^2 = 5.000000000000000?
A^3 + B^3 + C^3 = 4.000000000000000?
A^4 + B^4 + C^4 = 13.00000000000000?
A^5 + B^5 + C^5 = 16.00000000000000?

Same for the values $s,t,u$:

s, t, u = A*B, B*C, C*A

print "s + t + u    = %s" % (s+t+u)
print "st + tu + us = %s" % (s*t + t*u + u*s) 
print "s t u        = %s" % (s*t*u)

for k in [1..5]:
    print "s^%s + t^%s + u^%s = %s" % (k, k, k, s^k+t^k+u^k)

This gives:

s + t + u    = -2.000000000000000?
st + tu + us = -1.000000000000000?
s t u        = 1.000000000000000?
s^1 + t^1 + u^1 = -2.000000000000000?
s^2 + t^2 + u^2 = 6.000000000000000?
s^3 + t^3 + u^3 = -11.00000000000000?
s^4 + t^4 + u^4 = 26.00000000000000?
s^5 + t^5 + u^5 = -57.00000000000000?

Answer 2

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 $1$.

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