A little long recursion

Im pretty surprised that after 2 hours of huge audience there isn't a single solution. I solved this quite a while ago so the solution does not have the small details

Let $f(n)={ a }^{ n }+{ b }^{ n }+{ c }^{ n }$ and $a$ , $b$ and $c$ be the roots of ${ x }^{ 3 }+m{ x }^{ 2 }+nx+p$

Using newton sums,

$f(1)=-m\\ f(2)={ m }^{ 2 }-2n=-3\\ f(3)=-{ m }^{ 3 }+3mn-3p=-46\\ f(4)={ m }^{ 4 }-4{ m }^{ 2 }n+4mp+2{ n }^{ 2 }=-123$

It is pretty simple, now, just solve for $m$ and the answer would be $-m$ . Yeah... simple.

Using numerous substitutions, we get a Polynomial of degree 4

$\frac { { m }^{ 4 } }{ 6 } +3{ m }^{ 2 }+\frac { 184m }{ 3 } +127.5=0$

So, solving this tediously, (I used Vieta's formula), we get $2$ real and $2$ complex solutions:

$m=-5$

$m=-2.48411...$

$m=3.74206...-6.89843...i$

$m=3.74206...+6.89843... i$

Since $-m$ has to be an integer (stated), $f(1)=5$

Now, using newton sums again,

$f(1)={ S }_{ 1 }=5\\ f(2)={ S }_{ 1 }^{ 2 }-2{ S }_{ 2 }=-3\\ f(3)={ S }_{ 1 }f(2)-{ S }_{ 2 }f(1)+{ 3S }_{ 3 }=-46\\ f(n)=f(1)f(n-1)-{ S }_{ 2 }f(n-2)+{ S }_{ 3 }f(n-3)$

We can again tediously calculate that $f(10)=-104838$

Therefore, the answer is $\boxed{162}$

Thanks Alan Enrique Ontiveros Salazar! This question's rating has dropped by 58% since it's peak.

Read Julian Poon's solution on how to get that $a,b,c$ are roots of

$0=m^4+18m^2+368m+765=(m+5)(m^3-5m^2+43m+153)$

It's easy to see that $0=m^3-5m^2+43m+153$ has no integer roots. However, it has at least one real root $x$ and two complex roots $y$ and $z$ . Since $x+y+z=5$ from Vieta's formulas, and none of $a,b,c$ are integers, none of $x+y$ , $y+z$ , or $z+x$ are integers. If we add the other root $-5$ to any of these, they still won't be integers. Therefore, the only possible combination of these roots such that they sum to an integer is the roots of $0=m^3-5m^2+43m+153$ . This means that $a,b,c$ are the roots of $0=m^3-5m^2+43m+153$ .

Now apply Newton's sums $10$ times on the roots of this cubic equation to get the desired answer. There is no need to actually find the roots explicitly.