Consider the cubic polynomial $\begin{equation*} P(z)=z^{3}+Az^{2}+Bz+C \end{equation*}\tag{1},$ where the coefficients $A,B$ and $C$ are real numbers. If we denote its roots by $z_{1},z_{2}$ and $z_{3}$, then it factors as $\begin{eqnarray*} P(z) &=&\left( z-z_{1}\right) \left( z-z_{2}\right) (z-z_{3}) \\ &=&z^{3}-\left( z_{1}+z_{3}+z_{2}\right) z^{2}+\left( z_{1}z_{2}+z_{2}z_{3}+z_{1}z_{3}\right) z-z_{1}z_{2}z_{3}. \end{eqnarray*}\tag{2}$ The constant term is $\begin{equation*} P(0)=C=-z_{1}z_{2}z_{3} \end{equation*}.$ In the present case $A=-(b+6)$, $B=8b^{2}$ and $C=-7+b^{2}$. Since $z_{1}=1+i$ is a given solution, then $z_{2}=\overline{z}_{1}=\overline{1+i}=1-i$ is another solution, as you concluded. We thus have $z_{1}z_{2}=\left( 1+i\right) \left( 1-i\right)=2$ and $\begin{equation*} -7+b^{2}=-2z_{3} \end{equation*}\tag{3},$ whose solution is
$\begin{equation*} z_{3}=\frac{7-b^{2}}{2}. \end{equation*}\tag{4}$
Since $P(z_1)=P(z_2)=0$, we have $\begin{eqnarray*} &&\left( 1+i\right) ^{3}-(b+6)\left( 1+i\right) ^{2}+8b^{2}\left( 1+i\right) -7+b^{2} \\ &=&-9+9b^{2}+i\left( -10-2b+8b^{2}\right)=0, \end{eqnarray*}\tag{5}$
$\begin{eqnarray*} &&\left( 1-i\right) ^{3}-(b+6)\left( 1-i\right) ^{2}+8b^{2}\left( 1-i\right) -7+b^{2} \\ &=&-9+9b^{2}+i\left( 10+2b-8b^{2}\right)=0, \end{eqnarray*}\tag{6}$
which means that $b$ satisfies the system $\begin{equation*} \left\{ \begin{array}{c} -9+9b^{2}=0 \\ 10+2b-8b^{2}=0. \end{array} \right. \end{equation*}\tag{7}$ The solution of $(7)$ is $b=-1$. Using $(4)$ we find $z_{3}=3.\tag{8}$