The $\infty$ -radical madness 2

Let $A=\sqrt{8-b}$ , $b=\sqrt{8-c}$ and $c=\sqrt{8+A}$ . Then $A^2=8-b$ , $b^2=8-c$ and $c^2=8+A$ . Let $A=-a$ , then we have the following system: $\begin{aligned} a^2 &= 8-b & \cdots (1)\\ b^2 &= 8-c & \cdots (2)\\ c^2 &= 8-a & \cdots (3) \end{aligned}$ Let $m=a+b+c$ , $n=ab+bc+ca$ and $p=abc$ . Sum (1), (2) and (3) to obtain: $\begin{aligned} a^2+b^2+c^2 &= 24-(a+b+c)\\ m^2-2n &= 24-m\\ n &= \dfrac{1}{2}m^2+\dfrac{1}{2}m-12 & \cdots (4) \end{aligned}$ Subtract (1)-(2), (2)-(3) and (3)-(1): $\begin{aligned} a^2-b^2 &= -(b-c) &\implies (a+b)(a-b) &= -(b-c) & \cdots (5)\\ b^2-c^2 &= -(c-a) &\implies (b+c)(b-c) &= -(c-a) & \cdots (6)\\ c^2-a^2 &= -(a-b) &\implies (c+a)(c-a) &= -(a-b) & \cdots (7) \end{aligned}$ Multiply (5), (6) and (7), and remember that $a$ , $b$ and $c$ are all distinct: $\begin{aligned} (a+b)(a-b)(b+c)(b-c)(c+a)(c-a) & =-(a-b)(b-c)(c-a)\\ (a+b)(b+c)(c+a) &= -1\\ mn-p &= -1\\ p &= mn+1 & \cdots (8) \end{aligned}$ We also have, using (1), (2) and (3): $\begin{aligned} b(8-b^2)=(8-a^2)c &\implies 8b-b^3=8c-ca^2 & \cdots (9)\\ c(8-c^2)=(8-b^2)a &\implies 8c-c^3=8a-ab^2 & \cdots (10)\\ a(8-a^2)=(8-c^2)b &\implies 8a-a^3=8b-bc^2 & \cdots (11)\\ \end{aligned}$ Sum (9), (10) and (11): $\begin{aligned} a^3+b^3+c^3 &= ab^2+bc^2+ca^2\\ m^3-3mn+3p &= ab^2+bc^2+ca^2 & \cdots (12) \end{aligned}$ Multiply (1) by $c$ , (2) by $a$ and (3) by $b$ : $\begin{aligned} bc &= 8c-ca^2 & \cdots (13)\\ ca &= 8a-ab^2 & \cdots (14)\\ ab &= 8b-bc^2 & \cdots (15) \end{aligned}$ Sum (13), (14) and (15): $\begin{aligned} ab+bc+ca &= 8(a+b+c)-(ab^2+bc^2+ca^2)\\ ab^2+bc^2+ca^2 &= 8m-n & \cdots (16) \end{aligned}$ Equate (12) and (16): $\begin{aligned} m^3-3mn+3p &= 8m-n & \cdots (17) \end{aligned}$ Substitute (8) in (17): $\begin{aligned} m^3-3mn+3(mn+1) &= 8m-n\\ m^3+3 &= 8m-n & \cdots (18) \end{aligned}$ Substitute (4) in (18): $\begin{aligned} m^3+3 &= 8m-\left(\dfrac{1}{2}m^2+\dfrac{1}{2}m-12\right)\\ m^3+\dfrac{1}{2}m^2-\dfrac{15}{2}m-9 &= 0\\ (m-3)(m+2)\left(m+\dfrac{3}{2}\right) &= 0 & \cdots (19) \end{aligned}$ From (19), (4) and (8) we obtain the solutions $(m,n,p)=(3,-6,-17)$ , $(m,n,p)=(-2,-11,23)$ and $(m,n,p)=\left(-\dfrac{3}{2},-\dfrac{93}{8},\dfrac{295}{16}\right)$ . Since $a<0$ and $b,c>0$ , then $p=abc<0$ , so the only solution is $(m,n,p)=(3,-6,-17)$ .

Let $P(z)$ have roots $a,b,c$ . Then $P(z)=z^3-3z^2-6z+17$ , so we have to solve $z^3-3z^2-6z+17=0$ . Let $z=z'+1$ , then the equation becomes ${z'}^3-9{z'}+9=0$ . Since $a=-A$ we have $\begin{aligned} (-A-1)^3-9(-A-1)+9 &= 0 & \cdots (20) \end{aligned}$

To find $B$ notice that $B=\sqrt[3]{9+9B}$ , then: $\begin{aligned} B^3-9B-9 &= 0 & \cdots (21) \end{aligned}$ Sum (20) and (21): $\begin{aligned} (-A-1)^3+B^3-9(-A+B-1) &= 0\\ (-A+B-1)((-A-1)^2-(-A-1)B+B^2)-9(-A+B-1) &= 0\\ (-A+B-1)(A^2+2A+1+AB+B+B^2-9) &= 0 \end{aligned}$ Since the second factor cannot be zero (left as an exercise), we must have $-A+B-1=0$ , or $\boxed{B-A=1}$ .