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$z_1 = 1+i$ is a given solution.

I guess what I have to find is $z_2$ and $z_3$ in

$(z - (1 + i))(z - z_2)(z-z_3) = z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2$.

I tried to divide the polynomial by $(z - (1 + i))$, but that didn’t seem to work because of the $b$. According to the Complex conjugate root theorem $z_2 = \overline{z_1} = 1 - i$ is a solution too and somebody mentioned that it’s a hint that all coefficients are real. But I still don’t know how to proceed. What am I missing?

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If $1+i$ is a root of your cubic, then $1-i$ ought to be a root of your cubic as well if it is to have real coefficients. This means your cubic can be written as $$(z-(1+i))(z-(1-i))(z-x)=(z-x)(z^2-2z+2)$$ Expand out and compare coefficients. –  J. M. May 13 '12 at 8:35
    
More explicitly, consider $$\frac{z^3-(b+6)z^2+8b^2 z+b^2-7}{z^2-2z+2}=z-b-4+\frac{(b+1)(2(4b-5)z+b+1)}{z^2-2 z+2}$$... now, what $b$ can I use so that nasty rational expression at the far right zeroes out? –  J. M. May 13 '12 at 8:48
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If $z_{1}=1+i$ and $z_{2}=1-i$, since the independent term is $% -7+b^{2}=-\left( 1+i\right) \left( 1-i\right) z_{3}=-2z_{3}$ it would imply that $z_{3}=\frac{7-b^{2}}{2}$. However there is no $b$ such that the given cubic $z^{3}-(b+6)z^{2}+8b^{3}z-7+b^{2}=0$ thas these 3 solutions. Is there any typo? –  Américo Tavares May 13 '12 at 10:19
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It’s $b^2$ instead of $b^3$. –  mcb May 13 '12 at 10:27
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@mcb Thanks! I posted an answer. –  Américo Tavares May 13 '12 at 11:04
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2 Answers

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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 $$z_{3}=3.\tag{8}$$

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You have two unknowns here, the third root which we call $r$, and the variable $b$ in your equation. You can find two equations in these two unknowns to solve for both $r$ and $b$.

First, the $z^2$ coefficient is always $-1$ times the sum of the roots, so you have $$b + 6 = (1 + i) + (1 - i) + r = 2 + r$$ This reduces to $r = b + 4$. Next, the the constant term is equal to $(-1)^3$ times the product of the roots, so you have $$b^2 - 7 = -(1 + i)(1-i)r = -2r$$ You now can solve these two equations for $b$ and $r$. I substitute $r = b + 4$ in the last equation, and get $$b^2 - 7 = -2b - 8$$ This reduces to the quadratic equation $b^2 + 2b + 1 = 0$, which is solved only by $b = -1$. Then $r = b+ 4 = 3$ is your remaining root.

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Vieta is a nice thing... –  J. M. May 13 '12 at 14:15
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