Are they all close to power s of 2?

Let $P_1 = a + b + c$ , $P_2 = ab+bc+ca$ and $P_3 = abc$ . Using Vieta's formulas, we have $P_1 = -3$ , $P_2=4$ and $P_3 = 8$ .

Now, using Newton sums method, we have:

$\begin{aligned} S_1 & = P_1 = -3 \\ S_2 & = P_1S_1 - 2P_2 = (-3)(-3) - 2(4) = 1 \\ S_3 & = P_1S_2 - P_2S_1 +3P_3 = (-3)(1) - 4(-3) + 3(8) = 33 = 2^5 + 1 \\ S_4 & = P_1S_3 - P_2S_2 +P_3S_1 = (-3)(33) - 4(1)+ 8(-3) = -127 \\ S_5 & = P_1S_4 - P_2S_3 +P_3S_2 = (-3)(-127) - 4(33)+ 8(1) = 257 = 2^8+1 \end{aligned}$

$\Rightarrow m + n = 5 + 8 = \boxed{13}$

This cubic polynomial has one real root and 2 imaginary roots.

By observing the coefficients, we can observe that the real root is 1.

By using this, we can reduce the cubic to quadratic.

I used horner's method. (https://en.wikipedia.org/wiki/Horner%27s_method).

We can also take (x-1) common and get the quadratic.

So, the resultant quadratic is ${ x }^{ 2 }+4x+8$ . It is of the form,( $l{ x }^{ 2 }+mx+n$ . where l =1, m=4, n=8.

In the above quadratic, determinant\(m^{ 2 }-4ln) is less than zero, So, the roots are complex.

Now the main key points are sum of roots and product of roots. Let us consider 1 as a. Then the remaining roots are b and c.

b+c = -4 and bc = 8. Using this we can find S(3) and S(5).

S(3) :- .

\({ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }={ { 1 }^{ 3 }+(b+c) }^{ 3 }-3ab(a+b)\quad =\quad 1-64+96\quad =\quad 1+{ 2 }^{ m=5 }\).

S(5):-.

${ a }^{ 5 }+{ b }^{ 5 }+{ c }^{ 5 }={ 1 }^{ 5 }+({ b }^{ 3 }+{ c }^{ 3 })({ b }^{ 2 }+{ c }^{ 2 })-{ b }^{ 2 }{ c }^{ 2 }(b+c)=1+32\times 0-(-256)$ .

= $1+{ 2 }^{ n=8 }$ .

Thus, m=5 and n=8 and m+n=13.

Note:- $({ b }^{ 2 }+{ c }^{ 2 })={ (b+c) }^{ 2 }-2bc\quad =\quad 16-16\quad =\quad 0.$ .

its easy to note

$a + b + c = -3$

$ab + bc + ca = 4$

$abc = 8$

we would get

$a^2 + b^2 + c^2 = 1$

and then

$a^3 + b^3 + c^3 = 33$

thus m = 5

and using long calculation

$a^5 + b^5 + c^5 = 257$

thus n = 8