Experimental... again

If $\omega = -\tfrac12 + \tfrac12i\sqrt{3}$ is a primitive cube root of unity, consider the complex numbers $r \; = \; u + v \hspace{1cm} s \; = \; u\omega + v\omega^2 \hspace{1cm} t \; = \; u\omega^2 + v\omega$ Then we calculate $r+s+t = 0 \hspace{1cm} rs + rt + st \; = \; -3uv \hspace{1cm} rst \; = \; u^3 + v^3$ and hence $r,s,t$ will be the roots of $10X^3 - 30X - 29 = 0$ provided that $uv = 1$ and $u^3 + v^3 = \tfrac{29}{10}$ , which means that $u^3,v^3$ are the roots of the quadratic $0 \; = \; X^2 - \tfrac{29}{10}X + 1 \; = \; \tfrac{1}{10}(5X - 2)(2X - 5)$ so that we have $u^3,v^3 = \tfrac52,\tfrac25$ . Thus we deduce that $\sqrt[3]{a} + \sqrt[3]{b}$ is a root of $10X^3 - 30X - 29 = 0$ when $a =u^3= \tfrac52$ , $b =v^3= \tfrac25$ . This is the only real root of the cubic.

If we put $y = \sqrt[5]{a}$ and $z = \sqrt[5]{b}$ , then (note that $yz=1$ and $y^5 + z^5 = a+b = \tfrac{29}{10}$ ) $\begin{aligned} (y+z)^3 & = \; (y^3 + z^3) + 3(y+z) \\ (y+z)^5 & = \; \tfrac{29}{10} + 5(y^3 + z^3) + 10(y+z) \end{aligned}$ so that $(y+z)^5 \; = \; \tfrac{29}{10} + 5\big((y+z)^3 - 3(y+z)\big) + 10(y+z)$ and hence that $10(y+z)^5 - 50(y+z)^3 + 50(y+z) - 29 \; = \; 0$ as stated.

If we put $Y = \sqrt[7]{a}$ and $Z = \sqrt[7]{b}$ , then (this time $YZ = 1$ and $Y^7 + Z^7 = \tfrac{29}{10}$ ) $\begin{aligned} (Y+Z)^3 & = \; (Y^3 + Z^3) + 3(Y+Z) \\ (Y+Z)^5 & = \; (Y^5 + Z^5) + 5(Y^3 + Z^3) +10(Y + Z) \\ (Y+Z)^7 & = \; \tfrac{29}{10} + 7(Y^5+Z^5) + 21(Y^3 + Z^3) + 35(Y+Z) \\ & = \; \tfrac{29}{10} + 7\big[(Y+Z)^5 - 5(Y^3+Z^3) - 10(Y+Z)\big] + 21(Y^3+Z^3) + 35(Y+Z) \\ & = \; \tfrac{29}{10} + 7(Y+Z)^5 - 14(Y^3+Z^3) - 35(Y+Z) \; = \; \tfrac{29}{10} + 7(Y+Z)^5 - 14\big[(Y+Z)^3 - 3(Y+Z)\big] - 35(Y+Z) \\ & = \; \tfrac{29}{10} + 7(Y+Z)^5 - 14(Y+Z)^3 + 7(Y+Z) \end{aligned}$ so that $10(Y+Z)^7 - 70(Y+Z)^5 + 140(Y+Z)^3 - 70(Y+Z) - 29 \; = \; 0$ Eisenstein's Irreducibility Criterion (using $p=2$ ) tells us that $29x^7 + 70x^5 - 140x^3 + 70x - 10$ is irreducible over $\mathbb{Q}$ , and hence $10x^7 - 70x^5 + 140x^3 - 70x - 29$ is the minimal polynomial of $Y+Z$ . Thus we deduce that $|n_1| + |n_2| + |n_3| + |n_4| + |n_5| \; = \; 10 + 70 + 140 + 70 + 29 \; = \; \boxed{319}$

It is worth noting that, with the assumption that $a,b$ are real, they are uniquely defined (to within ordering) by the fact that $\sqrt[3]{a} + \sqrt[3]{b}$ is a root of the irreducible cubic $10x^3 - 30x - 29$ (and, if they were not real, the rest of the question would not be uniquely defined). Thus we do not need to know about the minimal polynomial of $\sqrt[5]{a} + \sqrt[5]{b}$ .

Root is {5/2}^(1/7) + {2/5}^(1/7) and the polynomial is (Guessed and verified with WolframAlpha)

10x^7 - 70x^5 + 140x^3 + 70x - 29,

Answer=(10+70+140+70+29)=319