Algebra sequel

Let $S_1 = x+y+z$ , $S_2 = xy+yz+zx$ , $S_3 = xyz$ and $P_n = x^n+y^n+z^n$ , where $n$ is a positive integer. Using Newton's sums method as follows:

$\begin{aligned} P_1 & = x+y+z = 1 \\ P_2 & = S_1P_1 - 2S_2 = (1)(1)-2S_2 = 2 & \implies S_2 = - \frac 12 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = (1)(2)-\left(-\frac 12\right)(1)+3S_3 = 3 & \implies S_3 = - \frac 16 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = (1)(3)-\left(-\frac 12\right)(2)+ \left(\frac 16\right)(1) = \frac {25}6 \end{aligned}$

$\implies a \times b = 25 \times 6 = \boxed{150}$

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+xz)$

$\Leftrightarrow 1 = 2+2(xy+yz+xz) \Rightarrow xy+yz+xz = -\frac{1}{2} \quad \boxed{*}$

$x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2 +z^2 -(xy + yz + zx)$

From $\boxed{*}$ , we have:

$3-3xyz=1(2-(-\frac{1}{2})) \Rightarrow xyz = \frac{1}{6} \quad \boxed{**}$

We can get $x^4 + y^4 + z^4$ by doing:

$(x^3+y^3+z^3)(x+y+z) = \boxed{x^4 + y^4 + z^4} + xy(x^2+y^2)+yz(y^2+z^2) +xz(x^2+z^2)$

We need to add $xyz^2 + yzx^2+xzy^2 = xyz(x+y+z) = \frac{1}{6}$ , so that both side becomes:

$(x^3+y^3+z^3)(x+y+z)+\frac{1}{6} = x^4 + y^4 + z^4 + xy(x^2+y^2+z^2)+yz(x^2+y^2+z^2) +xz(x^2+y^2+z^2)$

$(3)(1) + \frac{1}{6}= x^4 + y^4 + z^4 + xy(2)+yz(2) +xz(2)$

$\frac{19}{6}= x^4 + y^4 + z^4 + 2(-\frac{1}{2})$

$\ x^4 + y^4 + z^4 = \frac{25}{6}$ with 25 and 6 coprime, hence, the answer is $\boxed{150}$