x^9-5x^6+3x^3+1=(x^3)^3-5(x^3)^2+3x^3+1 factorises: (x^3-1)((x^3)^2-4x^3-1)=(x-1)(x^2+x+1)((x^3)^2-4x^3-1). (Let y=x^3 and x^9-5x^6+3x^3+1 becomes y^3-5y^2+3y+1. y=1 is a root of this expression so y-1 is a factor, and this factor can then be divided among not cubic in y to get (y-1)(y^2-4y-1). Then x^3 can be substituted back and x^3-1 is factorised further as shown above.)
(x^3)^2-4x^3-1, complete the square: (x^3)^2-4x^3+4-5=(x^3-2)^2-5=(x^3-2-sqrt(5))(x^3-2+sqrt(5)); x^3=2+sqrt(5), so x=1.618, -0.618. (Consider ((1/2)+sqrt(5)/2)^3=(1/8)(1+3sqrt(5)+15+5sqrt(5))=(1/8)(16+8sqrt(5))=2+sqrt(5); ((1/2)-sqrt(5)/2)^3=(1/8)(1-3sqrt(5)+15-5sqrt(5))=(1/8)(16-8sqrt(5))=2-sqrt(5). So x=(1/2)(1+sqrt(5)).)
Therefore, the solution to 9-degree polynomial is 1, (1/2)(1+sqrt(5)) (real roots only).
x^3-12x^2-2x+1 has 3 zeroes: 0.2186, 12.1577, -0.3763.
a, b and c have not been defined for a^5+b^5+c^5, so let's assume a, b and c are the three zeroes of the first polynomial.
We have:
(1)^5+((1/32)(1+sqrt(5))^5+(1/32)(1-sqrt(5))^5=
1+(1/32)(176+80sqrt(5)+176-80sqrt(5))=1+352/32=1+11=12.
(1+sqrt(5))^5=1+5sqrt(5)+10*5+10*5sqrt(5)+5*5^2+5^2sqrt(5)=176+80sqrt(5).)
If we do the same for the cubic expression we get 265618 as the approximate result.